Conic Sections in Precalculus: Equations, Graphs and Examples

 

Conic Sections in Precalculus: Equations, Graphs, Examples and Practice Problems

Conic sections are among the most useful topics in precalculus because they connect algebraic equations with geometric shapes. Circles, parabolas, ellipses and hyperbolas may look like four different subjects, but they all come from one geometric idea: the intersection of a plane with a double cone.

The challenge is learning how to recognize each conic from its equation, convert general equations into standard form, identify important features and sketch the graph accurately.

A reliable method is Scan, Square, Sketch:

1. Scan the equation to identify the likely conic.

2. Square the variables by completing the square and rewriting the equation in standard form.

3. Sketch the graph using the centre or vertex, axis direction, radius, foci and asymptotes where appropriate.

These steps are useful in Precalculus, Algebra 2 and AP Precalculus.

Key Formulas for Conic Sections

Circle:

(x − h)² + (y − k)² = r²

Parabola with a vertical axis:

(x − h)² = 4p(y − k)

Parabola with a horizontal axis:

(y − k)² = 4p(x − h)

Ellipse with a horizontal major axis:

(x − h)²/a² + (y − k)²/b² = 1, where a > b > 0

Ellipse with a vertical major axis:

(x − h)²/b² + (y − k)²/a² = 1, where a > b > 0

Ellipse relationship:

c² = a² − b²

Hyperbola with a horizontal transverse axis:

(x − h)²/a² − (y − k)²/b² = 1

Hyperbola with a vertical transverse axis:

(y − k)²/a² − (x − h)²/b² = 1

Hyperbola relationship:

c² = a² + b²

For these equations, (h, k) represents the centre or vertex, depending on the conic. The values a, b, c, p and r describe important geometric features.

1. What Are Conic Sections?

A conic section is a curve formed when a plane intersects a double cone. Changing the angle and position of the plane produces different curves.

The four main non-degenerate conic sections are:

• Circle

• Parabola

• Ellipse

• Hyperbola

A circle is a special case of an ellipse.

Conic sections appear in satellite orbits, reflecting telescopes, radio dishes, architectural structures, navigation and many mathematical models.

Circle

A circle consists of all points in a plane that are the same distance from a fixed point called the centre.

That fixed distance is the radius.

Parabola

A parabola consists of points that are equally distant from a fixed point called the focus and a fixed line called the directrix.

A parabola has one vertex and one axis of symmetry.

Ellipse

An ellipse consists of points for which the sum of the distances from two fixed points, called the foci, is constant.

Planetary orbits are commonly modelled using ellipses, with suitable physical assumptions.

Hyperbola

A hyperbola consists of points for which the absolute difference between the distances from two fixed points, called the foci, is constant.

Its two branches approach diagonal lines called asymptotes.

2. How to Identify a Conic from Its Equation

Consider an equation of the form:

Ax² + Cy² + Dx + Ey + F = 0

This form contains no xy term.

The coefficients of x² and y² provide an initial clue about the conic.

• If A = C and both are nonzero with the same sign, the equation is circle-type.

• If A and C have the same sign but different magnitudes, the equation is ellipse-type.

• If exactly one of A and C is zero, the equation is parabola-type, provided the remaining equation represents a non-degenerate parabola.

• If A and C have opposite signs, the equation is hyperbola-type.

These are preliminary classifications. An equation may represent no real points, a single point, intersecting lines, parallel lines or another degenerate case. The coefficients alone do not guarantee that the graph is a real, non-degenerate conic.

For example:

x² + y² = 25

Both squared terms have equal positive coefficients, so the equation represents a circle.

x² + 4y² = 16

The squared terms have positive but unequal coefficients, so the equation represents an ellipse.

x² = 8y

Only one variable is squared, so the equation represents a parabola.

x² − y² = 1

The squared terms have opposite signs, so the equation represents a hyperbola.

3. Circles: Equation, Centre and Radius

The standard equation of a circle is:

(x − h)² + (y − k)² = r²

Here:

• Centre: (h, k)

• Radius: r

• Diameter: 2r

The signs inside the brackets are opposite to the coordinates of the centre.

For example:

(x − 3)² + (y + 2)² = 25

The centre is (3, −2), and the radius is √25 = 5.

The circle extends 5 units to the left, right, above and below its centre.

Worked Example 1: Convert a Circle Equation into Standard Form

Rewrite the following equation in standard form and identify the centre and radius.

x² + y² − 6x + 4y − 12 = 0

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

(x² − 6x) + (y² + 4y) = 12

Step 2: Complete the square for each group.

For x² − 6x, half of −6 is −3, and (−3)² = 9.

For y² + 4y, half of 4 is 2, and 2² = 4.

Add 9 and 4 to both sides:

(x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4

Step 3: Write each group as a perfect square.

(x − 3)² + (y + 2)² = 25

Therefore:

Centre: (3, −2)

Radius: 5

The equation is now in standard circle form.

4. Parabolas: Vertex, Focus and Directrix

A parabola can open upward, downward, left or right.

The direction depends on which variable is squared and the sign of the coefficient on the other side.

Vertical Parabola

Standard equation:

(x − h)² = 4p(y − k)

Important features:

• Vertex: (h, k)

• Focus: (h, k + p)

• Directrix: y = k − p

• Axis of symmetry: x = h

If p > 0, the parabola opens upward.

If p < 0, the parabola opens downward.

Horizontal Parabola

Standard equation:

(y − k)² = 4p(x − h)

Important features:

• Vertex: (h, k)

• Focus: (h + p, k)

• Directrix: x = h − p

• Axis of symmetry: y = k

If p > 0, the parabola opens to the right.

If p < 0, the parabola opens to the left.

The value of p is a signed distance from the vertex to the focus. The distance between the focus and the directrix is 2|p|.

Worked Example 2: Find the Features of a Parabola

Find the vertex, focus, directrix and opening direction of:

x² − 6x − 8y + 1 = 0

Step 1: Group the x-terms.

x² − 6x = 8y − 1

Step 2: Complete the square.

Half of −6 is −3, and (−3)² = 9.

Add 9 to both sides:

x² − 6x + 9 = 8y − 1 + 9

(x − 3)² = 8y + 8

Factor the right-hand side:

(x − 3)² = 8(y + 1)

Compare this with:

(x − h)² = 4p(y − k)

Therefore:

h = 3

k = −1

4p = 8, so p = 2.

The features are:

Vertex: (3, −1)

Focus: (3, −1 + 2) = (3, 1)

Directrix: y = −1 − 2 = −3

Axis of symmetry: x = 3

Because p is positive, the parabola opens upward.

5. Ellipses: Standard Form, Foci and Eccentricity

An ellipse has two axes of symmetry. Its longer axis is called the major axis, and its shorter axis is called the minor axis.

For a horizontal major axis, the standard equation is:

(x − h)²/a² + (y − k)²/b² = 1

For a vertical major axis, the standard equation is:

(x − h)²/b² + (y − k)²/a² = 1

In both forms, a > b > 0.

The centre is (h, k).

The major-axis length is 2a.

The minor-axis length is 2b.

The distance from the centre to either focus is c, where:

c² = a² − b²

The eccentricity of an ellipse is:

e = c/a

For a non-circular ellipse, 0 < e < 1. A circle has e = 0 because its two foci coincide at the centre.

A smaller eccentricity indicates a shape closer to a circle. A value closer to 1 indicates a more elongated ellipse.

Worked Example 3: Convert an Ellipse Equation into Standard Form

Rewrite the equation and identify the centre, vertices and foci.

9x² + 4y² − 18x + 16y − 11 = 0

Step 1: Group the variables.

9(x² − 2x) + 4(y² + 4y) = 11

Step 2: Complete the square inside each bracket.

For x² − 2x, add 1.

For y² + 4y, add 4.

Because the expressions are multiplied by 9 and 4, respectively, the total additions are 9 and 16.

9(x² − 2x + 1) + 4(y² + 4y + 4) = 11 + 9 + 16

Step 3: Simplify.

9(x − 1)² + 4(y + 2)² = 36

Divide every term by 36:

(x − 1)²/4 + (y + 2)²/9 = 1

The centre is:

(1, −2)

The larger denominator is 9, beneath the y-term. Therefore, the major axis is vertical.

From the standard form:

a² = 9, so a = 3

b² = 4, so b = 2

Calculate c:

c² = a² − b²

c² = 9 − 4 = 5

c = √5

The vertices are:

(1, −2 + 3) = (1, 1)

(1, −2 − 3) = (1, −5)

The co-vertices are:

(1 + 2, −2) = (3, −2)

(1 − 2, −2) = (−1, −2)

The foci are:

(1, −2 + √5)

(1, −2 − √5)

This example illustrates why completing the square is essential: the centre and axis direction become clear only after the equation is converted into standard form.

6. Hyperbolas: Branches, Foci and Asymptotes

A hyperbola has two separate branches. Unlike an ellipse, its standard equation contains a subtraction between the squared terms.

Horizontal Hyperbola

(x − h)²/a² − (y − k)²/b² = 1

Centre: (h, k)

Vertices: (h ± a, k)

Foci: (h ± c, k)

Relationship:

c² = a² + b²

Asymptotes:

y − k = ±(b/a)(x − h)

The branches open to the left and right.

Vertical Hyperbola

(y − k)²/a² − (x − h)²/b² = 1

Centre: (h, k)

Vertices: (h, k ± a)

Foci: (h, k ± c)

Relationship:

c² = a² + b²

Asymptotes:

y − k = ±(a/b)(x − h)

The branches open upward and downward.

Remember that a² is associated with the positive term in the standard equation. The positive term determines the direction in which the hyperbola opens.

Worked Example 4: Find the Asymptotes of a Hyperbola

Rewrite the equation in standard form and identify its centre, vertices and asymptotes.

16x² − 9y² − 64x − 18y − 89 = 0

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

16(x² − 4x) − 9(y² + 2y) = 89

Step 2: Complete the square.

For x² − 4x, add 4.

For y² + 2y, add 1.

Account for the coefficients:

16(x² − 4x + 4) − 9(y² + 2y + 1) = 89 + 64 − 9

Step 3: Simplify.

16(x − 2)² − 9(y + 1)² = 144

Divide by 144:

(x − 2)²/9 − (y + 1)²/16 = 1

The centre is:

(2, −1)

Since the positive x-term comes first, the hyperbola opens to the left and right.

From the equation:

a² = 9, so a = 3

b² = 16, so b = 4

Calculate c:

c² = a² + b²

c² = 9 + 16 = 25

c = 5

The vertices are:

(2 + 3, −1) = (5, −1)

(2 − 3, −1) = (−1, −1)

The foci are:

(2 + 5, −1) = (7, −1)

(2 − 5, −1) = (−3, −1)

For a horizontal hyperbola, the asymptotes are:

y − k = ±(b/a)(x − h)

Substitute h = 2, k = −1, a = 3 and b = 4:

y + 1 = ±(4/3)(x − 2)

These two lines pass through the centre and guide the sketch of the hyperbola's branches.

7. Completing the Square: The Skill Behind Most Conic Problems

Completing the square converts an expression such as x² + 6x into a perfect-square expression.

The general pattern is:

x² + Bx = (x + B/2)² − (B/2)²

For example:

x² + 8x = (x + 4)² − 16

Check by expanding:

(x + 4)² − 16 = x² + 8x + 16 − 16 = x² + 8x

When completing the square in a conic equation, remember three rules:

1. Group terms involving the same variable.

2. Take half the coefficient of the linear term and square it.

3. If the group has a coefficient outside the brackets, account for that coefficient when balancing the equation.

For example:

2(x² + 6x)

The number needed inside the bracket is 9 because half of 6 is 3 and 3² = 9.

Adding 9 inside the bracket increases the full expression by 18, not 9:

2(x² + 6x + 9) = 2(x + 3)²

This is a frequent source of errors in ellipse and hyperbola questions.

8. How to Sketch a Conic Section

A correct sketch begins with the standard equation, not guesswork.

Sketching a Circle

1. Identify the centre (h, k).

2. Calculate the radius r.

3. Mark the centre on the coordinate plane.

4. Mark points r units above, below, left and right of the centre.

5. Draw a smooth circle through those points.

Sketching a Parabola

1. Find the vertex (h, k).

2. Determine p from 4p.

3. Locate the focus and directrix.

4. Identify the opening direction from the sign of p.

5. Draw a symmetric curve through the vertex.

Sketching an Ellipse

1. Find the centre.

2. Identify which denominator is larger.

3. Determine the major and minor axes.

4. Mark the vertices and co-vertices.

5. Calculate the foci if required.

6. Draw a smooth closed oval through the vertices and co-vertices.

Sketching a Hyperbola

1. Find the centre.

2. Identify the positive squared term.

3. Determine a and b.

4. Draw a guide rectangle centred at (h, k), with horizontal half-width a and vertical half-height b for a horizontal hyperbola.

5. Draw the asymptotes through opposite corners of the rectangle.

6. Sketch the two branches opening in the direction of the positive term.

For a vertical hyperbola, the guide rectangle has vertical half-height a and horizontal half-width b.

Asymptotes guide the shape but are not part of the hyperbola itself.

9. What Happens When an Equation Contains an xy Term?

A more general conic equation is:

Ax² + Bxy + Cy² + Dx + Ey + F = 0

When B ≠ 0, the conic may be rotated relative to the coordinate axes.

The discriminant of the quadratic terms is:

Δ = B² − 4AC

For a non-degenerate real conic, the usual classification is:

• Δ < 0: ellipse-type

• Δ = 0: parabola-type

• Δ > 0: hyperbola-type

A circle is a special ellipse. Degenerate cases and equations with no real points require additional checks, so the discriminant alone does not establish that a real, non-degenerate conic exists.

A rotation of axes can remove the xy term. When A ≠ C, a rotation angle θ satisfies:

tan(2θ) = B/(A − C)

If A = C and B ≠ 0, the corresponding rotation can be represented by 2θ = 90° modulo 180°, so θ = 45° modulo 90°.

This topic may be treated as an extension in precalculus, depending on the course syllabus.

10. Common Conic Section Mistakes

Mistake 1: Reading the centre's signs incorrectly

In:

(x − 4)² + (y + 3)² = 49

The centre is (4, −3), not (−4, 3).

Mistake 2: Confusing the direction of a parabola

In:

(x − 2)² = −12(y + 1)

We have 4p = −12, so p = −3.

The parabola opens downward, not upward.

Mistake 3: Choosing the wrong major axis of an ellipse

The larger denominator identifies the major-axis direction.

In:

(x − 1)²/16 + (y + 2)²/4 = 1

The major axis is horizontal because 16 is beneath the x-term.

Mistake 4: Using the wrong relationship for c

For an ellipse:

c² = a² − b²

For a hyperbola:

c² = a² + b²

Confusing these relationships changes the foci.

Mistake 5: Reversing the hyperbola's asymptote slopes

For a horizontal hyperbola:

y − k = ±(b/a)(x − h)

For a vertical hyperbola:

y − k = ±(a/b)(x − h)

The fractions are different.

Mistake 6: Forgetting the effect of outside coefficients

When completing the square, adding a value inside a bracket may change the overall expression by more than that value if the bracket has a coefficient.

Mistake 7: Assuming every quadratic equation represents a normal conic

Some quadratic equations have no real graph or represent degenerate curves. Check the transformed equation before naming the final graph.

11. Conic Sections Practice Problems

Try these questions before checking the answers.

Question 1: Circle

Find the centre and radius:

(x + 2)² + (y − 5)² = 36

Question 2: Parabola

Find the vertex, focus and directrix:

(x − 1)² = 12(y + 2)

Question 3: Ellipse

Identify the centre and major-axis direction:

(x + 3)²/25 + (y − 1)²/9 = 1

Question 4: Hyperbola

Find the centre and opening direction:

(y − 4)²/16 − (x + 2)²/9 = 1

Question 5: Completing the Square

Convert the equation into standard form:

x² + y² + 8x − 2y + 8 = 0

Question 6: Ellipse Foci

For the ellipse below, calculate c and identify the foci:

(x − 2)²/25 + (y + 1)²/9 = 1

Question 7: Hyperbola Asymptotes

Find the asymptotes:

(x − 3)²/4 − (y + 2)²/9 = 1

Question 8: Identify the Conic

What type of conic is represented by the equation?

x² − 6y = 0

12. Practice Answers

Answer 1

Centre: (−2, 5)

Radius: 6

Answer 2

Compare with (x − h)² = 4p(y − k).

Vertex: (1, −2)

4p = 12, so p = 3.

Focus: (1, 1)

Directrix: y = −5

Answer 3

Centre: (−3, 1)

The major axis is horizontal because 25 is the larger denominator.

Answer 4

Centre: (−2, 4)

The positive y-term comes first, so the hyperbola opens upward and downward.

Answer 5

Group the terms:

(x² + 8x) + (y² − 2y) = −8

Complete the square:

(x² + 8x + 16) + (y² − 2y + 1) = −8 + 16 + 1

Therefore:

(x + 4)² + (y − 1)² = 9

Centre: (−4, 1)

Radius: 3

Answer 6

a² = 25 and b² = 9.

c² = a² − b² = 25 − 9 = 16.

Therefore c = 4.

The centre is (2, −1), and the major axis is horizontal.

Foci: (2 + 4, −1) and (2 − 4, −1)

Foci: (6, −1) and (−2, −1)

Answer 7

Here a² = 4, so a = 2, and b² = 9, so b = 3.

The hyperbola is horizontal.

Its asymptotes are:

y − k = ±(b/a)(x − h)

Substitute h = 3 and k = −2:

y + 2 = ±(3/2)(x − 3)

Answer 8

The equation is:

x² = 6y

This is a parabola opening upward, with vertex (0, 0).

13. Why Conic Sections Matter in Precalculus

Conic sections combine several important mathematical skills:

• Factoring and expanding algebraic expressions

• Completing the square

• Interpreting equations and graphs

• Working with coordinate geometry

• Understanding functions and symmetry

• Using quadratic equations in geometric settings

These skills are relevant to Algebra 2, Precalculus and AP Precalculus. Conic sections also provide useful background for calculus, analytic geometry and physics.

The exact depth of conic sections varies across curricula. Some courses emphasize standard equations and graphing, while others also examine foci, eccentricity, rotated conics and more advanced applications.

Final Review: Scan, Square, Sketch

When you encounter a conic section equation, follow the same three steps.

Scan: Examine the squared terms and their signs to identify the likely conic.

Square: Complete the square and convert the equation into standard form.

Sketch: Use the centre or vertex, radius, axes, foci and asymptotes to construct the graph.

The most important habit is to avoid memorizing an equation without understanding what its parts mean. Once you can convert a general quadratic equation into standard form, the geometry becomes much easier to interpret.