Analytic Geometry and Conic Sections

Lesson: Locus Definitions

In analytic geometry, a locus is the set of all points that satisfy a given condition. This idea is very important because many familiar graphs and conic sections can be described as loci.

For example, a circle can be defined as the set of all points in a plane that are a fixed distance from a single point. That sentence is a locus definition. Instead of thinking of a graph as just a shape, we describe it precisely using a condition that every point on the graph must satisfy.

Understanding locus definitions helps you do three important things:

• translate a geometric description into an equation,
• recognize what kind of figure an equation represents,
• connect algebra and geometry in the coordinate plane.

1. What does “locus” mean?

A locus is the complete set of points that satisfy a rule. There are two key parts in this idea:

• Every point on the figure must satisfy the condition.
• Every point that satisfies the condition must be on the figure.

This means a locus is not just some points that work. It is all the points that work.

For a point \\(P(x,y)\\), we often use the distance formula to turn a geometric condition into algebra:

This formula is one of the main tools for deriving locus equations.

2. Common locus definitions

Many important graphs come from simple distance conditions.

• Circle: the set of all points a fixed distance from a fixed point.
• Perpendicular bisector of a segment: the set of all points equidistant from two fixed points.
• Parabola: the set of all points equidistant from a fixed point and a fixed line.
• Ellipse: the set of all points for which the sum of the distances from two fixed points is constant.
• Hyperbola: the set of all points for which the absolute difference of the distances from two fixed points is constant.

These are precise mathematical definitions. They describe each figure as a locus.

3. Turning a locus description into an equation

To find an equation from a locus definition, follow a process like this:

1. Choose a general point on the locus, usually \\(P(x,y)\\).
2. Write the condition using distances or coordinates.
3. Simplify the equation.
4. If needed, complete the square to rewrite it in standard form.
5. Identify the graph and its key features.

This process links the geometry definition to the algebraic equation.

4. Locus and standard form

Sometimes the equation you get from a locus is not immediately in a recognizable form. Then you simplify and complete the square.

For example, if an equation becomes

you can rewrite it by completing the square:

Now it is clear that the locus is a circle with center \\((3,-4)\\) and radius \\(4\\).

This is why complete-the-square skills are useful in analytic geometry.

5. Important idea: “set of all points”

A common mistake is to think a locus definition describes only a few special points. But the locus includes every point that satisfies the condition.

For instance, if a point is equidistant from two fixed points, the locus is not just the midpoint. It is the entire perpendicular bisector of the segment joining those points.

6. Worked Example 1: Circle from a locus definition

Problem: Find the equation of the locus of points that are 5 units from the point \\((2,-1)\\).

Step 1: Use a general point.

Let \\(P(x,y)\\) be any point on the locus.

Step 2: Apply the distance condition.

Step 3: Square both sides.

Answer: The locus is a circle with equation

Interpretation: center \\((2,-1)\\), radius \\(5\\).

This example shows the basic locus definition of a circle.

7. Worked Example 2: Points equidistant from two fixed points

Problem: Find the locus of points equidistant from \\(A(1,2)\\) and \\(B(5,6)\\).

Step 1: Let \\(P(x,y)\\) be a point on the locus.

Since \\(P\\) is equidistant from \\(A\\) and \\(B\\),

Using the distance formula:

Step 2: Square both sides.

Step 3: Expand both sides.

Simplify:

Answer: The locus is

This is a straight line. In fact, it is the perpendicular bisector of segment \\(AB\\).

Why does that make sense? Any point on the perpendicular bisector of a segment is the same distance from both endpoints.

8. Worked Example 3: Parabola from a locus definition

Problem: Find the equation of the locus of points equidistant from the point \\((0,2)\\) and the line \\(y=-2\\).

This is the locus definition of a parabola.

Step 1: Let \\(P(x,y)\\) be any point on the locus.

The distance from \\(P(x,y)\\) to the focus \\((0,2)\\) is

The distance from \\(P(x,y)\\) to the horizontal line \\(y=-2\\) is

For points on this parabola, the graph lies above the directrix, so the distance is \\(y+2\\).

Step 2: Set the distances equal.

Step 3: Square both sides.

Expand:

Answer: The locus is

This is a parabola opening upward.

Its standard form is \\(x^2=4py\\), so here \\(4p=8\\), giving \\(p=2\\). That matches the focus \\((0,2)\\) and directrix \\(y=-2\\).

9. Worked Example 4: Locus leading to a circle in general form

Problem: Find the locus of points whose distance from \\((3,4)\\) is the same as their distance from the origin plus 1 is not a standard conic definition, so instead consider this clearer condition:

Find the equation of the locus of points 4 units from \\((3,-2)\\), then write it in both standard form and general form.

Step 1: Use the circle definition.

Square both sides:

This is already the standard form.

Step 2: Expand to get general form.

Answer:

• Standard form: \\((x-3)^2+(y+2)^2=16\\)
• General form: \\(x^2+y^2-6x+4y-3=0\\)

This example shows how the same locus can be written in different algebraic forms.

10. How locus definitions connect to conic sections

Conic sections are not just equations to memorize. Each one has a geometric meaning as a locus.

• Circle: fixed distance from one point.
• Parabola: equal distance from a focus and a directrix.
• Ellipse: constant sum of distances from two foci.
• Hyperbola: constant difference of distances from two foci.

When you know the locus definition, it becomes easier to understand why a graph has its shape and how its equation is formed.

11. Graphing from a locus definition

To sketch a graph from a locus definition, ask these questions:

1. What condition defines the points?
2. Does the condition describe equal distance, fixed distance, or a sum/difference of distances?
3. What basic figure does that suggest: line, circle, parabola, ellipse, or hyperbola?
4. Can I find important features such as center, radius, focus, directrix, or midpoint?
5. Can I write the equation in standard form to graph it accurately?

For example, if points are equidistant from two fixed points, the graph is a line. If points are a fixed distance from one point, the graph is a circle.

12. Common mistakes to avoid

• Forgetting “all points”: a locus is the complete set, not just one example point.
• Using the wrong distance formula: be careful with signs inside brackets.
• Not squaring carefully: when removing square roots, square both sides correctly.
• Missing absolute value for distance to a line: distance must be non-negative.
• Stopping too early: simplify fully so the graph can be recognized.

13. Quick check for understanding

• If a point moves so that it is always 6 units from \\((0,0)\\), what is the locus? A circle.
• If a point is always the same distance from \\((2,1)\\) and \\((2,9)\\), what is the locus? The perpendicular bisector of the segment joining those points.
• If a point is equidistant from a fixed point and a fixed line, what is the locus? A parabola.

14. Brief summary

A locus is the set of all points that satisfy a given condition. In coordinate geometry, we usually represent a general point by \\(P(x,y)\\), write the condition using distances, and simplify to get an equation.

Locus definitions help explain many important graphs: circles, lines such as perpendicular bisectors, and conic sections like parabolas, ellipses, and hyperbolas. Once the equation is found, completing the square can help rewrite it in standard form so the graph is easier to identify and sketch.

Circles in the Coordinate Plane are one of the most important conic sections in analytic geometry. In this lesson, you will learn how to recognize a circle from its equation, how to write the standard equation of a circle, and how to start with a more complicated quadratic equation and rewrite it to find the circle’s center and radius exactly.

A circle is the set of all points in a plane that are the same distance from one fixed point. That fixed point is called the center, and the fixed distance is the radius.

In the coordinate plane, this definition leads directly to an equation. If the center is \\( (h,k) \\) and any point on the circle is \\( (x,y) \\), then the distance from \\( (x,y) \\) to \\( (h,k) \\) must always equal the radius \\( r \\).

Using the distance formula, we get:

Now square both sides:

This is the standard equation of a circle.

Standard form:

• \\( (h,k) \\) is the center
• \\( r \\) is the radius
• \\( r > 0 \\)

This form is very useful because it tells you the center and radius immediately.

For example, if a circle has equation

then its center is \\( (3,-2) \\) and its radius is \\( 5 \\), because \\( r^2=25 \\).

Be careful with signs. In standard form, the center is \\( (h,k) \\), but the equation uses \\( (x-h) \\) and \\( (y-k) \\). This means:

• \\( (x-4)^2 \\) gives \\( h=4 \\)
• \\( (x+4)^2=(x-(-4))^2 \\) gives \\( h=-4 \\)
• \\( (y-7)^2 \\) gives \\( k=7 \\)
• \\( (y+7)^2=(y-(-7))^2 \\) gives \\( k=-7 \\)

So the signs inside the parentheses are the opposite of the coordinates of the center.

Graphing a circle from standard form is straightforward:

1. Identify the center \\( (h,k) \\).
2. Find the radius \\( r \\).
3. Plot the center.
4. Move \\( r \\) units up, down, left, and right from the center to get key points.
5. Sketch the circle through those points.

For instance, for

the center is \\( (-1,4) \\) and the radius is \\( 3 \\). Key points would be:

• \\( (-1+3,4)=(2,4) \\)
• \\( (-1-3,4)=(-4,4) \\)
• \\( (-1,4+3)=(-1,7) \\)
• \\( (-1,4-3)=(-1,1) \\)

Sometimes a circle is not given in standard form. Instead, it may appear in a general quadratic form such as

where the coefficients of \\( x^2 \\) and \\( y^2 \\) are equal and there is no \\( xy \\) term. In that case, the graph may still be a circle, but you must rewrite the equation by completing the square.

Why completing the square works: it changes expressions like \\( x^2+6x \\) into a perfect square trinomial:

To complete the square for a variable:

1. Take the coefficient of the linear term.
2. Divide it by 2.
3. Square the result.
4. Add that number to make a perfect square trinomial.

For example:

• \\( x^2+8x \\): half of 8 is 4, and \\( 4^2=16 \\), so \\( x^2+8x+16=(x+4)^2 \\)
• \\( y^2-10y \\): half of \\( -10 \\) is \\( -5 \\), and \\( (-5)^2=25 \\), so \\( y^2-10y+25=(y-5)^2 \\)

When rewriting a circle equation, group the \\( x \\)-terms together and the \\( y \\)-terms together, then complete the square for each group.

Worked Example 1: Identify the center and radius from standard form

Find the center and radius of

Step 1: Compare to the standard form

Here, \\( h=5 \\), \\( k=-1 \\), and \\( r^2=16 \\).

Step 2: Find the radius.

Answer:

• Center: \\( (5,-1) \\)
• Radius: \\( 4 \\)

Worked Example 2: Write the equation of a circle from its center and radius

Write the equation of the circle with center \\( (-2,3) \\) and radius \\( 6 \\).

Use the standard form:

Substitute \\( h=-2 \\), \\( k=3 \\), and \\( r=6 \\):

Answer:

Worked Example 3: Rewrite a general equation in standard form

Find the center and radius of the circle:

Step 1: Group the \\( x \\)-terms and \\( y \\)-terms.

Step 2: Move the constant to the other side.

Step 3: Complete the square for each group.

For \\( x^2-6x \\): half of \\( -6 \\) is \\( -3 \\), and \\( (-3)^2=9 \\).

For \\( y^2+8y \\): half of 8 is 4, and \\( 4^2=16 \\).

Add these values to both sides:

Step 4: Rewrite as perfect squares.

Step 5: Read the center and radius.

• Center: \\( (3,-4) \\)
• Radius: \\( 6 \\)

Worked Example 4: A more challenging equation

Rewrite the equation in standard form and determine whether it represents a real circle:

Step 1: Group the variable terms.

Step 2: Move the constant term.

Step 3: Complete the square.

For \\( x^2+4x \\): half of 4 is 2, and \\( 2^2=4 \\).

For \\( y^2-12y \\): half of \\( -12 \\) is \\( -6 \\), and \\( (-6)^2=36 \\).

Add both to each side:

This means

So the radius is \\( 0 \\). This is not a circle with visible size; it is a single point at \\( (-2,6) \\). In many school problems, a circle is expected to have \\( r>0 \\), so this is called a degenerate case.

Important observations about general equations:

• If the equation can be rewritten as \\( (x-h)^2+(y-k)^2=r^2 \\) with \\( r^2>0 \\), then it is a real circle.
• If \\( r^2=0 \\), the graph is a single point.
• If \\( r^2<0 \\), there is no real graph, because a square cannot add up to a negative number.

For example, if you ended with

there would be no real circle, since the left side is always at least 0, but the right side is negative.

How to tell whether a quadratic equation is a circle before completing the square:

• Both \\( x^2 \\) and \\( y^2 \\) appear
• The coefficients of \\( x^2 \\) and \\( y^2 \\) are the same
• There is no \\( xy \\) term

So an equation like

can still represent a circle. First divide everything by 2:

Then complete the square as usual.

Common mistakes to avoid

• Sign errors: In \\( (x+3)^2 \\), the center coordinate is \\( -3 \\), not 3.
• Forgetting to square the radius: The equation uses \\( r^2 \\), not \\( r \\).
• Adding to only one side: When completing the square, every number added on the left must also be added on the right.
• Using the wrong number to complete the square: Always take half the linear coefficient, then square it.
• Not checking the final value of \\( r^2 \\): Make sure it is positive for a real circle.

Quick method for graphing once standard form is found

1. Find the center.
2. Find the radius.
3. Plot the center.
4. Plot four main points \\( r \\) units left, right, up, and down.
5. Draw a smooth circular curve through them.

If needed, you can also test additional points, but the center and radius are usually enough for an accurate sketch.

Final Summary

A circle is the set of all points the same distance from a center. Its standard equation is

where \\( (h,k) \\) is the center and \\( r \\) is the radius. If a circle is given in general quadratic form, you can identify it by equal coefficients on \\( x^2 \\) and \\( y^2 \\) and no \\( xy \\) term, then use completing the square to rewrite it in standard form. Once in standard form, the center, exact radius, and graph become much easier to find.

The Parabola as a Conic Section

A parabola is one of the conic sections. In analytic geometry, it can be described in two connected ways:

• as the graph of a quadratic equation, and
• as the locus of points that are equally distant from a fixed point and a fixed line.

That fixed point is called the focus, and the fixed line is called the directrix. This geometric definition is what makes a parabola a conic section.

In this lesson, you will learn how to use the focus-directrix definition to graph a parabola, how the value  controls its shape and position, and how to find the latus rectum.

1. The locus definition of a parabola

A parabola is the set of all points in a plane that are the same distance from:

• a fixed point, the focus, and
• a fixed line, the directrix.

If a point on the parabola is \\(P(x,y)\\), then:

This equality is the key idea behind every parabola.

2. Standard forms of a parabola

There are two main standard forms, depending on whether the parabola opens up/down or left/right.

(a) Vertical parabola

If the vertex is \\((h,k)\\), then the standard form is

For this form:

• the vertex is \\((h,k)\\)
• the axis of symmetry is \\(x=h\\)
• the focus is \\((h,k+p)\\)
• the directrix is \\(y=k-p\\)

If \\(p>0\\), the parabola opens upward. If \\(p<0\\), it opens downward.

(b) Horizontal parabola

If the vertex is \\((h,k)\\), then the standard form is

For this form:

• the vertex is \\((h,k)\\)
• the axis of symmetry is \\(y=k\\)
• the focus is \\((h+p,k)\\)
• the directrix is \\(x=h-p\\)

If \\(p>0\\), the parabola opens to the right. If \\(p<0\\), it opens to the left.

3. What does the p-value mean?

The value \\(p\\) measures the directed distance from the vertex to the focus. It also measures the directed distance from the vertex to the directrix.

This means:

• larger \\(|p|\\) gives a wider parabola
• smaller \\(|p|\\) gives a narrower parabola
• the sign of \\(p\\) tells the direction of opening

Be careful: in the equation, the coefficient is \\(4p\\), not just \\(p\\).

For example, if

then \\(4p=12\\), so \\(p=3\\), not 12.

4. How the standard form comes from the focus-directrix definition

Let us derive the equation of a simple parabola with:

• vertex at \\((0,0)\\)
• focus at \\((0,p)\\)
• directrix \\(y=-p\\)

Take any point \\((x,y)\\) on the parabola. By definition, its distance from the focus equals its distance from the directrix.

Distance from \\((x,y)\\) to the focus \\((0,p)\\):

Distance from \\((x,y)\\) to the directrix \\(y=-p\\):

So,

Now square both sides:

This is the standard form of a parabola opening upward with vertex at the origin.

By shifting the graph to a vertex \\((h,k)\\), we get

Similarly, a horizontal parabola has standard form

5. The latus rectum

The latus rectum is a line segment that:

• passes through the focus, and
• is perpendicular to the axis of symmetry.

Its endpoints lie on the parabola.

For every parabola in standard form, the length of the latus rectum is

This is an important result to remember.

Endpoints of the latus rectum

For \\((x-h)^2=4p(y-k)\\):

• focus: \\((h,k+p)\\)
• latus rectum is horizontal
• endpoints: \\((h-2p,k+p)\\) and \\((h+2p,k+p)\\)

For \\((y-k)^2=4p(x-h)\\):

• focus: \\((h+p,k)\\)
• latus rectum is vertical
• endpoints: \\((h+p,k-2p)\\) and \\((h+p,k+2p)\\)

6. Steps for graphing a parabola from its equation

1. Write the equation in standard form.
2. Identify whether it is vertical or horizontal.
3. Read the vertex \\((h,k)\\).
4. Find \\(p\\) from \\(4p\\).
5. Find the focus and directrix.
6. Mark the axis of symmetry.
7. Use the latus rectum to get extra points and sketch the curve.

If the equation is not already in standard form, you may need to complete the square.

Worked Example 1: Graphing from standard form

Graph the parabola

Step 1: Compare with standard form

The equation matches

So, \\(4p=8\\), which gives

Step 2: Identify the main features

• vertex: \\((0,0)\\)
• opens upward because \\(p>0\\)
• focus: \\((0,2)\\)
• directrix: \\(y=-2\\)
• axis of symmetry: \\(x=0\\)

Step 3: Find the latus rectum

Length of latus rectum:

The latus rectum passes through the focus \\((0,2)\\). Its endpoints are

Conclusion: Plot the vertex, focus, directrix, and latus rectum endpoints, then draw a smooth U-shaped curve opening upward.

Worked Example 2: Graphing a shifted parabola

Graph the parabola

Step 1: Compare with standard form

This matches

So:

• \\(h=3\\)
• \\(k=-1\\)
• \\(4p=-12\\), so \\(p=-3\\)

Step 2: Main features

• vertex: \\((3,-1)\\)
• opens downward because \\(p<0\\)
• focus: \\((3,-1-3)=(3,-4)\\)
• directrix: \\(y=-1-(-3)=2\\)
• axis of symmetry: \\(x=3\\)

Step 3: Latus rectum

Length:

Since the focus is \\((3,-4)\\), the latus rectum is horizontal. Its endpoints are

So the endpoints are

Conclusion: Draw a downward-opening parabola with vertex at \\((3,-1)\\), symmetric about \\(x=3\\).

Worked Example 3: Complete the square, then graph

Write the equation in standard form and graph:

Step 1: Group the y-terms

Step 2: Complete the square

Take half of \\(-6\\), which is \\(-3\\), and square it:

Add 9 to both sides:

Factor the right side:

Step 3: Compare with standard form

This matches

So:

• \\(h=-1\\)
• \\(k=3\\)
• \\(4p=8\\), so \\(p=2\\)

Step 4: Main features

• vertex: \\((-1,3)\\)
• opens right
• focus: \\((-1+2,3)=(1,3)\\)
• directrix: \\(x=-1-2=-3\\)
• axis of symmetry: \\(y=3\\)

Step 5: Latus rectum

Length:

Since the focus is \\((1,3)\\), the latus rectum is vertical. Its endpoints are

Conclusion: This is a right-opening parabola with vertex \\((-1,3)\\).

Worked Example 4: Find the equation from focus and directrix

Find the equation of the parabola with focus \\((2,5)\\) and directrix \\(y=1\\).

Step 1: Determine the axis and vertex

The directrix is horizontal, so the axis is vertical through \\(x=2\\).

The vertex lies halfway between the focus and directrix. The focus has \\(y=5\\), and the directrix is \\(y=1\\), so the midpoint is

So the vertex is

Step 2: Find p

The distance from the vertex to the focus is

Step 3: Use standard form

Since the axis is vertical, use

Substitute \\((h,k)=(2,3)\\) and \\(p=2\\):

Answer:

7. Common mistakes to avoid

• Forgetting that the equation uses \\(4p\\): if \\(4p=16\\), then \\(p=4\\).
• Mixing up vertical and horizontal forms: if \\((x-h)^2\\) is on one side, the parabola is vertical; if \\((y-k)^2\\) is on one side, the parabola is horizontal.
• Using the wrong sign for the focus or directrix: the sign of \\(p\\) determines direction.
• Not completing the square correctly: always add the same value to both sides.
• Confusing latus rectum length with p: the length is \\(4|p|\\), not \\(|p|\\).

8. Quick reference

• Vertical parabola: \\((x-h)^2=4p(y-k)\\)
• Focus: \\((h,k+p)\\)
• Directrix: \\(y=k-p\\)
• Axis: \\(x=h\\)

• Horizontal parabola: \\((y-k)^2=4p(x-h)\\)
• Focus: \\((h+p,k)\\)
• Directrix: \\(x=h-p\\)
• Axis: \\(y=k\\)

• Latus rectum length: \\(4|p|\\)
• Vertex is always halfway between the focus and the directrix along the axis of symmetry.

Summary

A parabola is the set of all points that are equally distant from a focus and a directrix. Its standard forms are \\((x-h)^2=4p(y-k)\\) and \\((y-k)^2=4p(x-h)\\), where \\(p\\) tells the direction of opening and the distance from the vertex to the focus.

To graph a parabola, identify the vertex, find \\(p\\), locate the focus and directrix, and use the latus rectum for extra points. If needed, complete the square first so the equation is in standard form.

The Ellipse is one of the four main conic sections. In coordinate geometry, an ellipse is a closed curve that looks like a stretched circle.

An ellipse can be defined as the set of all points in a plane such that the sum of the distances from any point on the curve to two fixed points is constant. These fixed points are called the foci (plural of focus).

This lesson will show you how to identify an ellipse, write its standard form, locate its center, major and minor axes, vertices, co-vertices, and foci, and graph it correctly. We will also connect the equation to the ellipse's locus definition.

1. The Locus Definition of an Ellipse

Suppose two fixed points, called foci, are placed on a plane. If a point moves so that the sum of its distances to the two foci stays the same, then the path traced out is an ellipse.

If the foci are labeled \(F_1\) and \(F_2\), and \(P(x,y)\) is any point on the ellipse, then:

$$PF_1 + PF_2 = \text{constant}$$

This constant is always greater than the distance between the foci. If the two foci move closer together, the ellipse becomes more circle-like. If they move farther apart, the ellipse becomes more stretched.

This property is important because it explains why the ellipse has two centers of pull, unlike a circle, which has only one center.

2. Parts of an Ellipse

To graph an ellipse, you should know its important parts.

• Center: the midpoint of the ellipse
• Major axis: the longest axis of the ellipse
• Minor axis: the shortest axis of the ellipse
• Vertices: endpoints of the major axis
• Co-vertices: endpoints of the minor axis
• Foci: the fixed points used in the locus definition

If the ellipse is centered at \((h,k)\), then all of these points are found relative to \((h,k)\).

3. Standard Forms of an Ellipse

The standard form depends on whether the major axis is horizontal or vertical.

Case 1: Horizontal major axis

$$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \qquad \text{where } a^2>b^2$$

Here:

• Center: \((h,k)\)
• Major axis is horizontal
• Vertices: \((h\pm a, k)\)
• Co-vertices: \((h, k\pm b)\)
• Foci: \((h\pm c, k)\)

Case 2: Vertical major axis

$$\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \qquad \text{where } a^2>b^2$$

Here:

• Center: \((h,k)\)
• Major axis is vertical
• Vertices: \((h, k\pm a)\)
• Co-vertices: \((h\pm b, k)\)
• Foci: \((h, k\pm c)\)

In both cases, \(a\) is the distance from the center to a vertex, \(b\) is the distance from the center to a co-vertex, and \(c\) is the distance from the center to a focus.

These values are related by:

$$c^2 = a^2 - b^2$$

This formula is very important. It helps you find the foci once you know the lengths of the major and minor axes.

4. How to Tell Whether an Equation is an Ellipse

An equation represents an ellipse when:

• both \(x^2\) and \(y^2\) terms are present,
• the coefficients of \(x^2\) and \(y^2\) are positive and different,
• and the graph forms a closed curve.

For example:

$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$

is an ellipse centered at the origin.

But if the denominators are equal, the graph is a circle, not an ellipse.

5. How to Graph an Ellipse from Standard Form

1. Find the center \((h,k)\).
2. Identify which denominator is larger. That tells you the direction of the major axis.
3. Take square roots of the denominators to find \(a\) and \(b\).
4. Use \(c^2=a^2-b^2\) to find \(c\).
5. Plot the center, vertices, co-vertices, and foci.
6. Draw a smooth closed curve through the vertices and co-vertices.

Remember: the larger denominator is always under the variable in the direction of the major axis.

6. Worked Example 1: Ellipse Centered at the Origin

Graph the ellipse:

$$\frac{x^2}{36} + \frac{y^2}{16} = 1$$

Step 1: Identify the center.

Since the equation is in standard form and there are no shifts, the center is \((0,0)\).

Step 2: Find \(a^2\) and \(b^2\).

The larger denominator is \(36\), so:

$$a^2=36, \quad b^2=16$$

Thus:

$$a=6, \quad b=4$$

Because the larger denominator is under \(x^2\), the major axis is horizontal.

Step 3: Find the vertices and co-vertices.

Vertices:

$$(-6,0), \ (6,0)$$

Co-vertices:

$$(0,-4), \ (0,4)$$

Step 4: Find the foci.

Use:

$$c^2=a^2-b^2=36-16=20$$

So:

$$c=\sqrt{20}=2\sqrt{5}$$

Foci:

$$(-2\sqrt{5},0), \ (2\sqrt{5},0)$$

Conclusion: The ellipse is centered at the origin, stretched more in the horizontal direction, with major axis length \(2a=12\) and minor axis length \(2b=8\).

7. Worked Example 2: Ellipse with Center \((h,k)\)

Graph the ellipse:

$$\frac{(x-3)^2}{9} + \frac{(y+2)^2}{25} = 1$$

Step 1: Find the center.

The center is:

$$ (h,k)=(3,-2) $$

Step 2: Identify the major axis.

The larger denominator is \(25\), and it is under \((y+2)^2\). So the major axis is vertical.

That means:

$$a^2=25, \quad b^2=9$$

So:

$$a=5, \quad b=3$$

Step 3: Find vertices and co-vertices.

Since the major axis is vertical:

Vertices:

$$ (3, -2\pm 5) $$

So the vertices are:

$$ (3,3), \ (3,-7) $$

Co-vertices:

$$ (3\pm 3, -2) $$

So the co-vertices are:

$$ (6,-2), \ (0,-2) $$

Step 4: Find the foci.

$$c^2=a^2-b^2=25-9=16$$

$$c=4$$

Since the major axis is vertical, the foci are:

$$ (3,-2\pm 4) $$

So:

$$ (3,2), \ (3,-6) $$

Conclusion: This ellipse is centered at \((3,-2)\) and is taller than it is wide.

8. Worked Example 3: Writing an Ellipse in Standard Form by Completing the Square

Rewrite and graph the equation:

$$9x^2 + 4y^2 - 54x + 16y + 61 = 0$$

Step 1: Group x-terms and y-terms.

$$9x^2 - 54x + 4y^2 + 16y + 61 = 0$$

$$9(x^2 - 6x) + 4(y^2 + 4y) + 61 = 0$$

Step 2: Complete the square inside each group.

For \(x^2-6x\), add and subtract \(9\) because \((-6/2)^2=9\).

For \(y^2+4y\), add and subtract \(4\) because \((4/2)^2=4\).

$$9(x^2 - 6x + 9 - 9) + 4(y^2 + 4y + 4 - 4) + 61 = 0$$

$$9[(x-3)^2 - 9] + 4[(y+2)^2 - 4] + 61 = 0$$

Step 3: Simplify.

$$9(x-3)^2 - 81 + 4(y+2)^2 - 16 + 61 = 0$$

$$9(x-3)^2 + 4(y+2)^2 - 36 = 0$$

$$9(x-3)^2 + 4(y+2)^2 = 36$$

Step 4: Divide by 36.

$$\frac{(x-3)^2}{4} + \frac{(y+2)^2}{9} = 1$$

Step 5: Identify the ellipse parts.

Center:

$$ (3,-2) $$

The larger denominator is \(9\), under the \(y\)-term, so the major axis is vertical.

$$a^2=9, \quad b^2=4$$

$$a=3, \quad b=2$$

Find \(c\):

$$c^2=9-4=5$$

$$c=\sqrt{5}$$

Vertices:

$$ (3,-2\pm 3) = (3,1), (3,-5) $$

Co-vertices:

$$ (3\pm 2,-2) = (5,-2), (1,-2) $$

Foci:

$$ (3,-2\pm \sqrt{5}) $$

This is the full graphing information.

9. Verifying the Constant Sum of Distances Property

For an ellipse, the sum of the distances from any point on the ellipse to the two foci is always the same. That constant sum is:

$$2a$$

This means if you pick any point \(P\) on the ellipse, then:

$$PF_1 + PF_2 = 2a$$

For example, in the ellipse

$$\frac{x^2}{36} + \frac{y^2}{16} = 1$$

we found \(a=6\), so the constant sum must be:

$$2a=12$$

Take the point \((6,0)\), which is a vertex. The foci are \((\pm 2\sqrt{5},0)\).

The distances are:

$$|6-2\sqrt{5}| \quad \text{and} \quad |6+2\sqrt{5}|$$

Their sum is:

$$ (6-2\sqrt{5}) + (6+2\sqrt{5}) = 12 $$

This confirms the ellipse property for that point. The same total will happen for every point on the ellipse.

10. Common Mistakes to Avoid

• Mixing up \(a\) and \(b\): \(a\) is always tied to the larger denominator in standard form.
• Using the wrong focus formula: for ellipses, use $$c^2=a^2-b^2$$ not addition.
• Placing vertices on the wrong axis: vertices lie along the major axis, co-vertices along the minor axis.
• Forgetting the signs in the center: in \((x-h)^2\), the center x-coordinate is \(h\); in \((y-k)^2\), the center y-coordinate is \(k\).
• Skipping completing the square carefully: when a coefficient is factored out, be careful to complete the square inside the parentheses correctly.

11. Quick Strategy for Any Ellipse Problem

1. Put the equation in standard form.
2. Read the center from the equation.
3. Find \(a^2\) and \(b^2\) from the denominators.
4. Decide whether the major axis is horizontal or vertical.
5. Use \(c^2=a^2-b^2\) to find the foci.
6. Plot center, vertices, co-vertices, and foci.
7. Sketch the ellipse as a smooth closed curve.
8. If asked about the locus property, remember that the sum of distances to the foci is \(2a\).

12. Brief Summary

An ellipse is the set of all points whose distances from two fixed points add to a constant. Its graph can be described using the center, major axis, minor axis, vertices, co-vertices, and foci.

In standard form, the larger denominator shows the direction of the major axis. The relationship $$c^2=a^2-b^2$$ helps locate the foci, and the constant sum of distances from the foci is always $$2a$$.

Once you can rewrite equations into standard form and identify the important parts, graphing ellipses becomes a clear step-by-step process.

The Hyperbola is one of the conic sections. In Grade 12, you need to know how to recognize its equation, rewrite it into standard form, and then graph it using its key features: the center, transverse axis, conjugate axis, vertices, foci, and asymptotes.

A hyperbola looks like two separate curves that open away from each other. Unlike an ellipse, which is closed, a hyperbola has two branches. Its shape is guided by straight lines called asymptotes, which the branches get closer to but never touch.

This lesson will show you how to graph a hyperbola step by step, including how to complete the square when the equation is not already in standard form.

1. What is a hyperbola?

A hyperbola is the set of all points in a plane such that the difference of the distances from two fixed points is constant. These fixed points are called the foci.

For graphing, you usually work from its equation instead of the distance definition. The important idea is that a hyperbola has:

• a center,
• two vertices,
• two foci,
• a transverse axis along which the hyperbola opens,
• a conjugate axis perpendicular to the transverse axis,
• and two diagonal asymptotes.

2. Standard forms of a hyperbola

The standard form tells you whether the hyperbola opens left-right or up-down.

Horizontal transverse axis (opens left and right):

Its center is \((h,k)\).

The positive term is under the \(x\)-part, so the hyperbola opens horizontally.

Vertical transverse axis (opens up and down):

Its center is again \((h,k)\).

The positive term is under the \(y\)-part, so the hyperbola opens vertically.

Important note: In a hyperbola, one squared term is positive and the other is negative. That sign difference is a key way to recognize a hyperbola.

3. Meaning of \(a\), \(b\), and \(c\)

In the standard forms above:

• \(a\) measures the distance from the center to each vertex along the transverse axis.
• \(b\) helps determine the rectangle used to draw the asymptotes.
• \(c\) measures the distance from the center to each focus.

For hyperbolas, \(a\), \(b\), and \(c\) satisfy:

This is different from an ellipse, where the relationship is \(c^2=a^2-b^2\). For hyperbolas, you add.

4. Key parts of the graph

If the hyperbola is horizontal,

• Center: \((h,k)\)
• Vertices: \((h-a,k)\) and \((h+a,k)\)
• Foci: \((h-c,k)\) and \((h+c,k)\)
• Transverse axis: horizontal line through the center
• Conjugate axis: vertical line through the center
• Asymptotes: $$y-k=\pm \frac{b}{a}(x-h)$$

If the hyperbola is vertical,

• Center: \((h,k)\)
• Vertices: \((h,k-a)\) and \((h,k+a)\)
• Foci: \((h,k-c)\) and \((h,k+c)\)
• Transverse axis: vertical line through the center
• Conjugate axis: horizontal line through the center
• Asymptotes: $$y-k=\pm \frac{a}{b}(x-h)$$

5. How asymptotes are drawn

To sketch a hyperbola neatly, first imagine a rectangle centered at \((h,k)\).

For a horizontal hyperbola, the rectangle extends:

• \(a\) units left and right from the center,
• \(b\) units up and down from the center.

So the rectangle has width \(2a\) and height \(2b\). The asymptotes are the diagonals of this rectangle.

For a vertical hyperbola, the rectangle still uses \(a\) and \(b\), but now the hyperbola opens upward and downward. The asymptotes are again the diagonals of the rectangle.

This is why asymptotes are often called the defining diagonal asymptotes of the hyperbola.

6. Steps for graphing a hyperbola from standard form

1. Identify whether the equation is horizontal or vertical by checking which term is positive.
2. Find the center \((h,k)\).
3. Read off \(a^2\) and \(b^2\), then calculate \(a\) and \(b\).
4. Find the vertices using the direction of the transverse axis.
5. Use \(c^2=a^2+b^2\) to find \(c\), then locate the foci.
6. Draw the rectangle centered at \((h,k)\) using \(a\) and \(b\).
7. Draw the diagonal asymptotes.
8. Sketch the two branches opening toward the vertices and approaching the asymptotes.

7. Worked Example 1: Hyperbola centered at the origin

Graph the hyperbola

Step 1: Identify the form.

The \(x^2\)-term is positive, so this is a horizontal hyperbola.

Step 2: Find the center.

Since there are no shifts, the center is \((0,0)\).

Step 3: Find \(a\) and \(b\).

Here, \(a^2=9\) and \(b^2=4\), so

Step 4: Find the vertices.

Because it is horizontal, move left and right from the center by \(a=3\):

Step 5: Find the foci.

So the foci are

Step 6: Find the asymptotes.

For a horizontal hyperbola centered at the origin:

Final graph features:

• Center: \((0,0)\)
• Transverse axis: horizontal
• Conjugate axis: vertical
• Vertices: \((\pm 3,0)\)
• Foci: \((\pm \sqrt{13},0)\)
• Asymptotes: \(y=\pm \frac{2}{3}x\)

The branches open left and right, passing through the vertices and approaching the asymptotes.

8. Worked Example 2: Shifted vertical hyperbola

Graph the hyperbola

Step 1: Identify the form.

The \(y\)-term is positive, so the hyperbola is vertical.

Step 2: Find the center.

From \((y+1)^2=(y-(-1))^2\) and \((x-2)^2\), the center is

Step 3: Find \(a\) and \(b\).

Step 4: Find the vertices.

Since the transverse axis is vertical, move up and down from the center by \(a=4\):

Step 5: Find the foci.

So the foci are

Step 6: Find the asymptotes.

For a vertical hyperbola:

Substitute \((h,k)=(2,-1)\), \(a=4\), \(b=5\):

Final graph features:

• Center: \((2,-1)\)
• Transverse axis: vertical
• Conjugate axis: horizontal
• Vertices: \((2,-5)\) and \((2,3)\)
• Foci: \((2,-1\pm \sqrt{41})\)
• Asymptotes: \(y+1=\pm \frac{4}{5}(x-2)\)

The branches open upward and downward.

9. Completing the square to get standard form

Sometimes the equation of a hyperbola is not given in standard form. Then you must rearrange it and complete the square.

Worked Example 3: Rewrite and graph

Put the equation into standard form and identify its key features:

Step 1: Group the \(x\)-terms and \(y\)-terms.

Step 2: Complete the square inside each bracket.

For \(x^2-2x\), add and subtract \(1\).

For \(y^2-4y\), add and subtract \(4\).

Step 3: Expand and simplify.

Step 4: Divide by 4 to get standard form.

So the hyperbola is horizontal, with center \((1,2)\).

Step 5: Find \(a\), \(b\), and \(c\).

Step 6: Find vertices, foci, and asymptotes.

Vertices:

So the vertices are

Foci:

Asymptotes for a horizontal hyperbola:

Final graph features:

• Center: \((1,2)\)
• Transverse axis: horizontal
• Conjugate axis: vertical
• Vertices: \(\left(\frac{1}{3},2\right)\) and \(\left(\frac{5}{3},2\right)\)
• Foci: \(\left(1\pm \frac{\sqrt{13}}{3},2\right)\)
• Asymptotes: \(y-2=\pm \frac{3}{2}(x-1)\)

10. Worked Example 4: Find all graphing information from standard form

For the hyperbola

determine the transverse axis, conjugate axis, vertices, foci, and asymptotes.

Step 1: Orientation

The \(y\)-term is positive, so the hyperbola is vertical.

Step 2: Center

Step 3: Values of \(a\), \(b\), and \(c\)

Step 4: Transverse and conjugate axes

• Transverse axis: vertical line through the center, so it lies on \(x=-1\)
• Conjugate axis: horizontal line through the center, so it lies on \(y=4\)

Step 5: Vertices

Move up and down from the center by \(a=3\):

Step 6: Foci

Move up and down by \(c=5\):

Step 7: Asymptotes

For a vertical hyperbola:

Substitute \((h,k)=(-1,4)\):

Answer:

• Center: \((-1,4)\)
• Transverse axis: vertical, line \(x=-1\)
• Conjugate axis: horizontal, line \(y=4\)
• Vertices: \((-1,1)\) and \((-1,7)\)
• Foci: \((-1,-1)\) and \((-1,9)\)
• Asymptotes: \(y-4=\pm \frac{3}{4}(x+1)\)

11. Common mistakes to avoid

• Using the ellipse rule for \(c\): For a hyperbola, use \(c^2=a^2+b^2\), not subtraction.
• Choosing the wrong direction: The variable with the positive term tells you whether the transverse axis is horizontal or vertical.
• Mixing up \(a\) and \(b\): \(a\) is always linked to the positive term in standard form.
• Incorrect center: Be careful with signs. For example, \((x+3)^2\) means \(h=-3\), not \(3\).
• Wrong asymptote formula: Horizontal and vertical hyperbolas use different slopes.

12. Quick checklist for graphing a hyperbola

• Is one squared term positive and the other negative?
• What is the center \((h,k)\)?
• Is the transverse axis horizontal or vertical?
• What are \(a\), \(b\), and \(c\)?
• What are the vertices?
• What are the foci?
• What are the asymptotes?
• Did you sketch both branches in the correct direction?

13. Summary

A hyperbola is a conic section with two separate branches. Its equation has one positive squared term and one negative squared term. The variable in the positive term tells you the direction of the transverse axis.

To graph a hyperbola, find the center, identify whether it opens horizontally or vertically, calculate the vertices and foci, and draw the diagonal asymptotes. If the equation is not in standard form, complete the square first. With these steps, you can graph even more complex hyperbolas accurately in the Cartesian plane.

Conic Classification and the Discriminant

In analytic geometry, many curves can be described by a general second-degree equation in two variables:

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

Here, the coefficients \(A, B, C, D, E,\) and \(F\) are constants, and at least one of \(A, B, C\) must be nonzero. This equation can represent a circle, parabola, ellipse, or hyperbola. These are called conic sections, or simply conics.

A very powerful way to identify the type of conic is to use the discriminant:

$$B^2-4AC$$

This value tells us the conic type before we fully rewrite the equation into standard form. In this lesson, you will learn how to use this discriminant, how it connects to the shape of a conic, and how it works together with completing the square.

1. The General Second-Degree Equation

The equation

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

contains several kinds of terms:

• Quadratic terms: \(Ax^2\), \(Bxy\), \(Cy^2\)
• Linear terms: \(Dx\), \(Ey\)
• Constant term: \(F\)

The quadratic terms are the most important for classification. In particular, the values of \(A\), \(B\), and \(C\) determine the value of \(B^2-4AC\), and this tells us what kind of conic we have.

2. The Discriminant Rule

For the equation

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

compute the discriminant:

$$Delta=B^2-4AC$$

Then classify the conic using these rules:

• If \(B^2-4AC<0\), the conic is an ellipse or a circle.
• If \(B^2-4AC=0\), the conic is a parabola.
• If \(B^2-4AC>0\), the conic is a hyperbola.

A circle is actually a special kind of ellipse. When there is no \(xy\)-term and the coefficients of \(x^2\) and \(y^2\) are equal, the ellipse is a circle.

So if \(B=0\) and \(A=C\), then the conic may be a circle, depending on whether the equation gives real points.

3. Why the Discriminant Works

The discriminant compares the quadratic part of the equation:

$$Ax^2+Bxy+Cy^2$$

Different conics have different patterns in these squared terms:

• Parabola: only one variable is effectively squared in standard form.
• Ellipse: both variables are squared, and the squared terms have the same sign.
• Hyperbola: both variables are squared, but the squared terms have opposite signs.

The value of \(B^2-4AC\) captures this pattern even before you complete the square.

4. Important Notes About the \(xy\)-Term

If \(B\neq 0\), then the equation has an \(xy\)-term. This often means the conic is rotated in the plane.

Even when a conic is rotated, the discriminant still correctly classifies it. That is one reason why the discriminant is so useful.

For example, a rotated ellipse may not look like the usual horizontal or vertical ellipse equation at first, but \(B^2-4AC\) will still be negative.

5. Connection to Standard Forms

After classifying a conic, we often rewrite it into standard form by grouping terms and completing the square. Here are the common standard forms.

Circle:

$$ (x-h)^2+(y-k)^2=r^2 $$

Ellipse:

$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$

Hyperbola:

$$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1 \quad \text{or} \quad \frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1$$

Parabola:

$$ (x-h)^2=4p(y-k) \quad \text{or} \quad (y-k)^2=4p(x-h) $$

The discriminant helps you know which standard form to expect.

6. Quick Classification Guide

1. Identify \(A\), \(B\), and \(C\) from the equation \(Ax^2+Bxy+Cy^2+Dx+Ey+F=0\).
2. Compute \(B^2-4AC\).
3. Use the sign of the result:

• Negative  ellipse or circle
• Zero  parabola
• Positive  hyperbola

Then, if needed, rewrite the equation into standard form to find the center, vertex, radii, or axes.

Worked Example 1: Classify a Simple Conic

Classify the equation

$$3x^2+12x+y-5=0$$

Step 1: Identify \(A\), \(B\), and \(C\).

There is an \(x^2\)-term, but no \(xy\)-term and no \(y^2\)-term. So:

$$A=3, \quad B=0, \quad C=0$$

Step 2: Compute the discriminant.

$$B^2-4AC=0^2-4(3)(0)=0$$

Step 3: Classify.

Since \(B^2-4AC=0\), the conic is a parabola.

Step 4: Confirm by rewriting.

$$3x^2+12x+y-5=0$$

$$y=-3x^2-12x+5$$

Factor the coefficient of the \(x\)-terms:

$$y=-3(x^2+4x)+5$$

Complete the square inside:

$$y=-3(x^2+4x+4)+5+12$$

$$y=-3(x+2)^2+17$$

This is clearly a parabola.

Worked Example 2: Distinguish Circle from Ellipse

Classify the equation

$$4x^2+4y^2-16x+8y+7=0$$

Step 1: Identify \(A\), \(B\), and \(C\).

$$A=4, \quad B=0, \quad C=4$$

Step 2: Compute the discriminant.

$$B^2-4AC=0^2-4(4)(4)=-64$$

Since the discriminant is negative, the conic is an ellipse or circle.

Step 3: Check whether it is a circle.

Because \(B=0\) and \(A=C\), the squared terms have equal coefficients, so this suggests a circle.

Step 4: Rewrite in standard form.

$$4x^2+4y^2-16x+8y+7=0$$

Divide everything by 4:

$$x^2+y^2-4x+2y+\frac{7}{4}=0$$

Group the \(x\)-terms and \(y\)-terms:

$$ (x^2-4x)+(y^2+2y)=-\frac{7}{4} $$

Complete the square:

$$ (x^2-4x+4)+(y^2+2y+1)=-\frac{7}{4}+4+1 $$

$$ (x-2)^2+(y+1)^2=\frac{13}{4} $$

This is a circle with center \((2,-1)\) and radius \(\frac{\sqrt{13}}{2}\).

Worked Example 3: Classify a Hyperbola

Classify the equation

$$9x^2-4y^2+18x+16y-11=0$$

Step 1: Identify \(A\), \(B\), and \(C\).

$$A=9, \quad B=0, \quad C=-4$$

Step 2: Compute the discriminant.

$$B^2-4AC=0^2-4(9)(-4)=144$$

Since \(144>0\), the conic is a hyperbola.

Step 3: Rewrite in standard form.

$$9x^2-4y^2+18x+16y-11=0$$

Group terms:

$$ (9x^2+18x) + (-4y^2+16y) = 11 $$

Factor the leading coefficients from each group:

$$9(x^2+2x)-4(y^2-4y)=11$$

Complete the square in each bracket:

$$9(x^2+2x+1)-4(y^2-4y+4)=11+9-16$$

$$9(x+1)^2-4(y-2)^2=4$$

Divide by 4:

$$\frac{9(x+1)^2}{4}-\frac{4(y-2)^2}{4}=1$$

$$\frac{(x+1)^2}{4/9}-\frac{(y-2)^2}{1}=1$$

This is a hyperbola centered at \((-1,2)\).

Worked Example 4: A Rotated Conic with an \(xy\)-Term

Classify the equation

$$x^2+4xy+y^2-6=0$$

Step 1: Identify \(A\), \(B\), and \(C\).

$$A=1, \quad B=4, \quad C=1$$

Step 2: Compute the discriminant.

$$B^2-4AC=4^2-4(1)(1)=16-4=12$$

Step 3: Classify.

Since \(12>0\), the conic is a hyperbola.

This equation has an \(xy\)-term, so the hyperbola is rotated. Even though it is not in standard form, the discriminant still tells us the correct conic type immediately.

7. Common Mistakes to Avoid

• Forgetting missing terms: If there is no \(xy\)-term, then \(B=0\). If there is no \(y^2\)-term, then \(C=0\).
• Using the wrong coefficients: Make sure you match the equation exactly to \(Ax^2+Bxy+Cy^2+Dx+Ey+F=0\).
• Confusing ellipse and circle: A circle happens when the squared terms have equal coefficients and there is no \(xy\)-term.
• Ignoring whether the graph is real: Sometimes an equation may classify like an ellipse or circle, but after rewriting, it may not produce real points. In most classification questions, the discriminant still identifies the conic type from the quadratic equation form.

8. How This Helps with Graphing

When you are asked to graph a conic from a general equation, the discriminant gives you a quick first step. It tells you what you are looking for before you begin algebraic work.

Then you can:

• complete the square,
• rewrite into standard form,
• find the center or vertex,
• identify the direction of opening or major axes,
• and sketch the graph accurately.

This is especially useful on longer problems, where recognizing the conic early saves time and prevents mistakes.

9. Final Summary

The general second-degree equation of a conic is

$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$

To classify the conic, use the discriminant

$$B^2-4AC$$

• If \(B^2-4AC<0\), the conic is an ellipse or circle.
• If \(B^2-4AC=0\), the conic is a parabola.
• If \(B^2-4AC>0\), the conic is a hyperbola.

After classification, completing the square helps rewrite the equation into standard form so you can graph it and identify important features. The discriminant is a quick and reliable tool, even when the equation includes an \(xy\)-term and the conic is rotated.

Translation and Rotation of Axes are methods used in analytic geometry to make equations easier to understand and graph.

When we study conic sections such as circles, parabolas, ellipses, and hyperbolas, their equations are often easiest to recognize in a simple form. But many equations are given in a more complicated form, with linear terms like \(x\) and \(y\), or a mixed term like \(xy\).

By translating the axes, we shift the origin to a new point. This helps remove linear terms.

By rotating the axes, we turn the coordinate grid through an angle. This helps remove the mixed term \(xy\).

These transformations do not change the actual graph. They only change the coordinate system we use to describe it.

Why do we do this?

• To rewrite a conic in a simpler form
• To identify the type of conic more easily
• To find the center, vertex, axes, and orientation of the graph
• To make sketching and analysis easier

1. General second-degree equation

Many conics can be written in the general form

The terms have different roles:

• \(Ax^2, Bxy, Cy^2\): quadratic terms
• \(Dx, Ey\): linear terms
• \(F\): constant term

If the equation contains:

• linear terms, we often use translation
• an \(xy\) term, we often use rotation
• both, we may need both translation and rotation

2. Translation of axes

Suppose we move the origin from \((0,0)\) to a new point \((h,k)\). Let the new coordinates be \((X,Y)\).

The relationship between old and new coordinates is:

Equivalently,

This means:

• \(X\) measures horizontal distance from the new origin
• \(Y\) measures vertical distance from the new origin

We choose \((h,k)\) so that the equation becomes simpler, usually by removing linear terms.

How translation helps

If a conic is centered at \((h,k)\), then shifting the origin to \((h,k)\) often gives a standard form.

For example, the circle

looks complicated at first, but after completing the square we can find its center. That same center can be used for translation.

Completing the square review

To complete the square:

• For \(x^2 + bx\), add and subtract \(\left(\frac{b}{2}\right)^2\)
• For \(y^2 + cy\), add and subtract \(\left(\frac{c}{2}\right)^2\)

Example:

3. Rotation of axes

Sometimes an equation has a mixed term \(xy\), which means the conic is tilted relative to the usual \(x\)- and \(y\)-axes.

To remove the \(xy\) term, we rotate the axes by an angle \(\theta\). Let the new coordinates be \((X,Y)\). Then:

These formulas describe a counterclockwise rotation of the axes by angle \(\theta\).

After substitution, the equation is rewritten in terms of \(X\) and \(Y\). We choose \(\theta\) so that the \(XY\) term disappears.

Key formula for eliminating the \(xy\) term

For the equation

the angle \(\theta\) that removes the mixed term satisfies

This formula is very important.

Notes about this formula

• If \(B=0\), there is no mixed term, so rotation may not be needed.
• If \(A=C\), then \(\tan 2\theta\) is undefined, which means \(2\theta = 90^\circ\), so \(\theta = 45^\circ\).
• The purpose is not to memorize every expanded expression, but to know how to choose the angle and simplify the conic.

4. When to translate and when to rotate

1. Start with the general equation.
2. If there is an \(xy\) term, consider rotation first.
3. If there are linear terms, consider translation.
4. After simplifying, rewrite the equation in standard form.
5. Identify the conic and graph it.

In many school-level problems, translation and rotation are studied separately first, then together in more advanced examples.

5. Worked Example 1: Translation to remove linear terms

Simplify the equation

Step 1: Group the \(x\)-terms and \(y\)-terms

Step 2: Complete the square

Substitute these into the equation:

This is a circle with center \((2,-3)\) and radius \(5\).

Step 3: Translate the axes

Choose the new origin at \((2,-3)\). Then

Substitute into the standard form:

Result: In the new coordinate system, the circle is centered at the origin and has the simple equation \(X^2 + Y^2 = 25\).

What we learn: Translation shifts the graph description so that the conic becomes easier to recognize.

6. Worked Example 2: Translation for a parabola

Simplify and identify the conic:

Step 1: Complete the square in \(y\)

Substitute:

This is a parabola of the form

So the vertex is \((-2,3)\), and \(4a=4\), so \(a=1\). It opens to the right.

Step 2: Translate the axes

Move the origin to the vertex \((-2,3)\). Then

Or equivalently, \(X = x+2\), \(Y = y-3\).

Substitute into the simplified form:

Result: In the new axes, the parabola has the standard equation \(Y^2 = 4X\).

What we learn: Translation is especially useful when the vertex or center is not at the origin.

7. Worked Example 3: Rotation to remove the \(xy\) term

Simplify the equation

This equation has a mixed term \(4xy\), so we use rotation.

Here,

Use the formula

The denominator is \(0\), so \(\tan 2\theta\) is undefined. Therefore,

Now use

Substitute into the equation:

First compute each part.

Now substitute:

Combine the first and third terms:

This is a hyperbola in the rotated axes.

Result: After rotation by \(45^\circ\), the equation becomes

What we learn: The conic was tilted in the original axes, but in the rotated axes it becomes a standard hyperbola.

8. Worked Example 4: Rotation and then identification

Consider the equation

This contains only a mixed term, so it suggests a rotated conic.

Rewrite in general form:

So \(A=0\), \(B=1\), \(C=0\).

Use

So again, \(\theta = 45^\circ\).

Use

Substitute into \(xy=2\):

This is a standard hyperbola in the rotated coordinate system.

Result: The graph of \(xy=2\) is a hyperbola whose axes are rotated \(45^\circ\) relative to the usual coordinate axes.

9. Important ideas for graphing conics after transformation

• Translation changes the location of the center or vertex, but not the shape.
• Rotation changes the orientation of the conic, but not the type of conic.
• After simplifying the equation, graph it in the new axes first.
• Then interpret that graph back in the original axes.

10. Quick guide: how to decide what to do

• If you see only \(x^2\) and \(y^2\) with linear terms, try completing the square and translation.
• If you see an \(xy\) term, try rotation.
• If the equation still has linear terms after rotation, then use translation next.
• Always rewrite the final equation in a standard form if possible.

11. Common mistakes to avoid

• Mixing up old and new coordinates: If the new origin is \((h,k)\), then \(x=X+h\) and \(y=Y+k\).
• Using the wrong sign in translation: Be careful when the new origin has negative coordinates.
• Forgetting the purpose of rotation: Rotation is mainly used to remove the \(xy\) term.
• Incorrectly completing the square: Always add and subtract the same value.
• Stopping too early: After transforming the equation, identify the conic and describe its key features.

12. Final summary

Translation and rotation of axes are tools that help us simplify conic equations.

Translation shifts the origin to a new point and is useful for removing linear terms or moving the center/vertex to the origin. The basic formulas are

Rotation turns the coordinate axes by an angle \(\theta\) and is useful for removing the mixed term \(xy\). The formulas are

The angle for removing the mixed term is found using

Once the equation is simplified, we can identify the conic more easily and graph it in standard form.

Conics in Polar Coordinates gives us one powerful way to describe circles, ellipses, parabolas, and hyperbolas using the same idea. Instead of starting with equations in the Cartesian plane like \(x^2+y^2=r^2\), we describe a conic by how each point is related to a focus and a directrix.

This lesson shows how all conics can be represented by a single polar equation. You will learn what the focus, directrix, and eccentricity mean, how the polar equation is derived, and how to identify which conic is being represented.

In polar coordinates, a point is written as \((r,\theta)\), where \(r\) is the distance from the pole and \(\theta\) is the angle from the polar axis. For conics in polar form, we usually place the focus at the pole. That makes the equation much simpler and reveals the shared structure of all conic sections.

1. The locus definition of a conic

A conic is the set of all points \(P\) such that the distance from \(P\) to a fixed point called the focus has a constant ratio to the distance from \(P\) to a fixed line called the directrix.

This constant ratio is called the eccentricity, written \(e\). So the defining condition is

Since the focus is placed at the pole in polar coordinates, the distance from a point \((r,\theta)\) to the focus is simply \(r\).

The value of \(e\) tells us which conic we have:

• Circle: a special case related to \(e=0\)
• Ellipse: \(0<e<1\)
• Parabola: \(e=1\)
• Hyperbola: \(e>1\)

The bigger the eccentricity, the more stretched the conic becomes.

2. Deriving the polar equation of a conic

Let the focus be at the pole. Suppose the directrix is the vertical line

where \(d>0\). A point \(P(r,\theta)\) has Cartesian coordinate \(x=r\cos\theta\). Its distance to the directrix \(x=d\) is

assuming the point lies on the same side where this quantity is positive.

Using the conic definition,

Now solve for \(r\):

This is one standard polar form of a conic.

If instead the directrix is

then the distance from \((r,\theta)\) to the directrix is

and the equation becomes

Similarly, if the directrix is horizontal, such as \(y=d\) or \(y=-d\), then \(\sin\theta\) appears instead of \(\cos\theta\).

So the common polar forms are

3. A useful simplified form

It is common to write

where

The quantity \(\ell\) is called the semi-latus rectum. For Grade 12, the most important idea is that \(\ell\) is just the constant in the numerator, and it helps determine the size of the conic.

4. How the equation tells you the conic type

In the equation

the eccentricity still controls the conic type:

• If \(0<e<1\), the graph is an ellipse.
• If \(e=1\), the graph is a parabola.
• If \(e>1\), the graph is a hyperbola.

The \(\cos\theta\) or \(\sin\theta\) tells the orientation of the conic.

• \(\cos\theta\) usually means the conic opens or stretches left-right.
• \(\sin\theta\) usually means the conic opens or stretches up-down.

The sign also matters:

• \(1+e\cos\theta\) and \(1-e\cos\theta\) correspond to directrices on opposite sides.
• \(1+e\sin\theta\) and \(1-e\sin\theta\) do the same for horizontal directrices.

5. Important graphing ideas

When graphing a conic in polar form, remember these key facts:

1. The focus is at the pole.
2. The value of \(e\) tells you the conic type.
3. The denominator shows how \(r\) changes with angle.
4. Values of \(\theta\) that make the denominator close to zero can make \(r\) very large.

For example, if \(e>1\), then there may be angles where the denominator becomes zero. That signals branches of a hyperbola extending far away.

For a parabola, the denominator becomes zero at exactly one important direction, matching the open shape of the curve.

6. Connecting polar and Cartesian thinking

You already know conics in Cartesian form, such as:

• Ellipse: \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)
• Parabola: \(y^2=4ax\)
• Hyperbola: \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

Polar form does not replace these equations. Instead, it gives another way to describe the same curves. Polar form is especially helpful when the focus is important, because the focus is built directly into the equation.

To move between systems, use

These relationships allow you to rewrite a polar equation in Cartesian form if needed.

Worked Example 1: Identify the conic

Determine the conic represented by

Step 1: Identify the eccentricity.

Compare with the standard form \(r=\frac{\ell}{1+e\cos\theta}\). Here,

Step 2: Classify the conic.

Since \(0<e<1\), the graph is an ellipse.

Step 3: Note the orientation.

Because the equation contains \(\cos\theta\), the conic is oriented along the horizontal direction.

Answer: The conic is an ellipse with focus at the pole.

Worked Example 2: Find the directrix

Given

find the eccentricity and directrix.

Step 1: Compare with standard form.

We compare with

So

Step 2: Solve for \(d\).

Step 3: Interpret the directrix.

The form \(1-e\cos\theta\) corresponds to the directrix

Therefore, the directrix is

Step 4: Classify the conic.

Since \(e=2>1\), the conic is a hyperbola.

Answer: Eccentricity \(e=2\), directrix \(x=-4\), and the conic is a hyperbola.

Worked Example 3: Derive the polar equation of a parabola

A conic has focus at the pole, directrix \(x=3\), and eccentricity \(e=1\). Find its polar equation.

Step 1: Use the formula for directrix \(x=d\).

Substitute \(e=1\) and \(d=3\):

Step 2: Classify the conic.

Since \(e=1\), this is a parabola.

Answer:

Worked Example 4: Convert to Cartesian form

Convert the polar equation

into a Cartesian equation.

Step 1: Multiply both sides by the denominator.

Step 2: Expand.

Step 3: Use \(x=r\cos\theta\).

Step 4: Replace \(r\) with \(\sqrt{x^2+y^2}\).

Step 5: Solve if desired.

Square both sides:

This is a parabola in Cartesian form.

Answer:

7. Common mistakes to avoid

• Confusing \(e\) with the numerator. The eccentricity is the number multiplying \(\cos\theta\) or \(\sin\theta\), not the whole numerator.
• Forgetting the focus is at the pole. In these standard polar forms, the focus is at the origin of the polar system.
• Mixing up the sign and directrix. Be careful whether the denominator is \(1+e\cos\theta\) or \(1-e\cos\theta\).
• Misclassifying the conic. Always classify from the value of \(e\): less than 1, equal to 1, or greater than 1.
• Using Cartesian ideas too early. First identify the conic from the polar form before converting.

8. Quick reference table

9. Final summary

All conic sections can be described by one polar model based on a focus, a directrix, and an eccentricity. When the focus is placed at the pole, the standard forms become

The value of \(e\) identifies the conic:

• \(0<e<1\): ellipse
• \(e=1\): parabola
• \(e>1\): hyperbola

By recognizing the eccentricity, the numerator, and whether the equation uses \(\cos\theta\) or \(\sin\theta\), you can identify, interpret, and graph conics in polar coordinates with confidence.