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Chapter 8: Conic Sections

Precalculus Textbook • Estimated reading time: 30 min

Chapter Contents

  1. 8.1 Introduction to Conic Sections
  2. 8.2 Parabolas
  3. 8.3 Ellipses
  4. 8.4 Hyperbolas
  5. 8.5 Identifying Conics from the General Equation
  6. 8.6 Rotation of Axes (Advanced)
  7. 8.7 Applications of Conic Sections
  8. 8.8 Interactive Conic Explorer
  9. 8.9 Practice Problems

Conic sections are among the most elegant and widely applicable curves in mathematics. They arise naturally when a plane intersects a double-napped cone at different angles, producing four distinct curves: circles, parabolas, ellipses, and hyperbolas. These curves appear throughout science and engineering, from the orbits of planets to the design of satellite dishes and suspension bridges. In this chapter, we develop each conic section algebraically, derive its key geometric properties, and explore real-world applications.

8.1 Introduction to Conic Sections

Imagine a double-napped right circular cone extending infinitely in both directions. When a plane slices through this cone, the intersection produces a curve called a conic section. The type of curve depends on the angle at which the plane meets the cone:

Definition: Conic Section

A conic section (or simply conic) is the set of all points $P$ in the plane such that the ratio of the distance from $P$ to a fixed point $F$ (called the focus) to the distance from $P$ to a fixed line $\ell$ (called the directrix) is a constant $e$ called the eccentricity:

$$\frac{d(P, F)}{d(P, \ell)} = e$$

When $e = 1$, the conic is a parabola. When $0 < e < 1$, it is an ellipse. When $e > 1$, it is a hyperbola. A circle is the special case $e = 0$.

This unified definition shows that all conic sections belong to a single family of curves, distinguished only by their eccentricity. Throughout this chapter, we will study each type individually, deriving standard forms and key properties.

8.2 Parabolas

A parabola is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix). This reflective property is the reason parabolic mirrors and satellite dishes can focus incoming signals to a single point.

Definition: Parabola

A parabola is the set of all points $(x, y)$ in the plane that are equidistant from a fixed point $F$ (the focus) and a fixed line $\ell$ (the directrix). The midpoint between the focus and the directrix is called the vertex. The line through the focus perpendicular to the directrix is the axis of symmetry.

Standard Forms of a Parabola

When the vertex is at the origin, there are four orientations:

Equation Opens Focus Directrix
$y^2 = 4px$ Right ($p > 0$) $(p, 0)$ $x = -p$
$y^2 = 4px$ Left ($p < 0$) $(p, 0)$ $x = -p$
$x^2 = 4py$ Up ($p > 0$) $(0, p)$ $y = -p$
$x^2 = 4py$ Down ($p < 0$) $(0, p)$ $y = -p$

When the vertex is translated to $(h, k)$, the equations become:

Reflective Property of the Parabola

Any ray traveling parallel to the axis of symmetry of a parabola will reflect off the surface and pass through the focus. Conversely, any ray emanating from the focus will reflect off the parabola and travel parallel to the axis. This is why parabolic mirrors concentrate light at the focal point.

Example 8.2.1 — Finding the Focus and Directrix

Find the focus, directrix, and vertex of the parabola $x^2 = -12y$.

Solution. This is in the form $x^2 = 4py$. Comparing, we have $4p = -12$, so $p = -3$.

Since $p < 0$, the parabola opens downward.

Example 8.2.2 — Writing the Equation of a Parabola

Write the equation of the parabola with vertex $(2, -1)$ and focus $(2, 3)$.

Solution. The vertex and focus share the same $x$-coordinate, so the axis of symmetry is vertical. We have $h = 2$, $k = -1$, and the focus is at $(h, k + p) = (2, 3)$.

Solving: $k + p = 3 \implies -1 + p = 3 \implies p = 4$.

Using the vertical-axis form:

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

The directrix is $y = k - p = -1 - 4 = -5$.

Example 8.2.3 — Completing the Square

Convert $y^2 + 6y + 8x + 1 = 0$ to standard form and identify the vertex, focus, and directrix.

Solution. Group the $y$-terms and complete the square:

$$(y^2 + 6y + 9) = -8x - 1 + 9$$ $$(y + 3)^2 = -8x + 8 = -8(x - 1)$$

This is in the form $(y - k)^2 = 4p(x - h)$ with $h = 1$, $k = -3$, and $4p = -8$, so $p = -2$.

The parabola opens to the left.

8.3 Ellipses

An ellipse looks like a stretched or compressed circle. It is defined by two focal points, and the sum of the distances from any point on the ellipse to the two foci is constant. This property makes ellipses fundamental to orbital mechanics: planets, moons, and satellites all follow elliptical paths.

Definition: Ellipse

An ellipse is the set of all points $(x, y)$ in the plane such that the sum of the distances from $(x, y)$ to two fixed points $F_1$ and $F_2$ (the foci) is a constant $2a$:

$$d(P, F_1) + d(P, F_2) = 2a$$

The center is the midpoint of the foci. The major axis has length $2a$ and the minor axis has length $2b$, where $a > b > 0$. The relationship between $a$, $b$, and the focal distance $c$ is $c^2 = a^2 - b^2$.

Standard Forms of an Ellipse (Center at Origin)

Form Major Axis Vertices Foci
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, $a > b$ Horizontal $(\pm a, 0)$ $(\pm c, 0)$
$\dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1$, $a > b$ Vertical $(0, \pm a)$ $(0, \pm c)$

When the center is translated to $(h, k)$, simply replace $x$ with $(x - h)$ and $y$ with $(y - k)$.

Eccentricity of an Ellipse

Eccentricity

The eccentricity of an ellipse is $e = \dfrac{c}{a}$, where $0 < e < 1$. When $e$ is close to $0$, the ellipse is nearly circular. When $e$ is close to $1$, the ellipse is highly elongated. A circle corresponds to $e = 0$ (when $c = 0$, meaning the two foci coincide).

Example 8.3.1 — Graphing an Ellipse

Sketch the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$ and find the foci, vertices, and eccentricity.

Solution. Here $a^2 = 25$ and $b^2 = 9$, so $a = 5$ and $b = 3$. Since $a^2$ is under $x^2$, the major axis is horizontal.

Example 8.3.2 — Translated Ellipse

Find the center, foci, and vertices of $\dfrac{(x - 3)^2}{16} + \dfrac{(y + 2)^2}{49} = 1$.

Solution. The center is $(h, k) = (3, -2)$. Here $a^2 = 49$ (under the $y$-term) and $b^2 = 16$, so $a = 7$ and $b = 4$. The major axis is vertical.

Example 8.3.3 — Writing an Ellipse Equation from Conditions

Find the equation of the ellipse with foci $(\pm 3, 0)$ and a major axis of length $10$.

Solution. The foci are on the $x$-axis, so the major axis is horizontal. We have $c = 3$ and $2a = 10$, so $a = 5$.

Using $c^2 = a^2 - b^2$: $9 = 25 - b^2 \implies b^2 = 16$.

$$\frac{x^2}{25} + \frac{y^2}{16} = 1$$

8.4 Hyperbolas

A hyperbola consists of two separate branches that open in opposite directions. While an ellipse is defined by a constant sum of distances to the foci, a hyperbola is defined by a constant difference. Hyperbolas appear in applications ranging from LORAN navigation to the cooling tower profiles of nuclear power plants.

Definition: Hyperbola

A hyperbola is the set of all points $(x, y)$ in the plane such that the absolute value of the difference of the distances from $(x, y)$ to two fixed points $F_1$ and $F_2$ (the foci) is a constant $2a$:

$$|d(P, F_1) - d(P, F_2)| = 2a$$

The center is the midpoint of the foci. The transverse axis has length $2a$ and connects the vertices. The conjugate axis has length $2b$. The relationship is $c^2 = a^2 + b^2$.

Standard Forms of a Hyperbola (Center at Origin)

Form Opens Vertices Foci Asymptotes
$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ Left/Right $(\pm a, 0)$ $(\pm c, 0)$ $y = \pm \dfrac{b}{a}x$
$\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ Up/Down $(0, \pm a)$ $(0, \pm c)$ $y = \pm \dfrac{a}{b}x$
Key Distinction: In the hyperbola equation, $a^2$ is always the denominator under the positive term. Unlike an ellipse, $a$ does not need to be greater than $b$ for a hyperbola. For the foci, always use $c^2 = a^2 + b^2$ (addition, not subtraction).

Asymptotes

As the branches of a hyperbola extend outward, they approach (but never touch) two straight lines called asymptotes. These asymptotes pass through the center and form a guide box of dimensions $2a \times 2b$ centered at the origin. The diagonals of this box are the asymptotes.

Example 8.4.1 — Analyzing a Hyperbola

Find the vertices, foci, and asymptotes of $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.

Solution. Here $a^2 = 9$ and $b^2 = 16$, so $a = 3$ and $b = 4$. The positive term contains $x^2$, so the hyperbola opens left and right.

Example 8.4.2 — Completing the Square for a Hyperbola

Rewrite $4x^2 - 9y^2 + 8x + 36y - 68 = 0$ in standard form.

Solution. Group and factor:

$$4(x^2 + 2x) - 9(y^2 - 4y) = 68$$

Complete the square in each group:

$$4(x^2 + 2x + 1) - 9(y^2 - 4y + 4) = 68 + 4 - 36 = 36$$ $$4(x + 1)^2 - 9(y - 2)^2 = 36$$

Divide by 36:

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

Center: $(-1, 2)$, $a = 3$, $b = 2$, $c = \sqrt{13}$. The hyperbola opens left and right with asymptotes $y - 2 = \pm \dfrac{2}{3}(x + 1)$.

Example 8.4.3 — Writing a Hyperbola Equation

Find the equation of the hyperbola with vertices $(0, \pm 4)$ and asymptotes $y = \pm 2x$.

Solution. The vertices are on the $y$-axis, so the transverse axis is vertical. The form is $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ with $a = 4$.

For a vertical transverse axis, the asymptotes are $y = \pm \dfrac{a}{b}x$. Setting $\dfrac{a}{b} = 2$: $\dfrac{4}{b} = 2 \implies b = 2$.

$$\frac{y^2}{16} - \frac{x^2}{4} = 1$$

8.5 Identifying Conics from the General Equation

Every conic section can be written in the general second-degree form:

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

When there is no $xy$-term (that is, $B = 0$), identifying the conic is straightforward by examining the coefficients $A$ and $C$:

Discriminant Test (when $B = 0$)

Given $Ax^2 + Cy^2 + Dx + Ey + F = 0$ with $A$ and $C$ not both zero:

When $B \neq 0$, the conic is rotated. The nature of the conic is determined by the discriminant $B^2 - 4AC$:

General Discriminant Test

For the general equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$:

(Degenerate cases such as a point, a line, or two intersecting lines are possible.)

Example 8.5.1 — Classifying Conics

Classify each conic without completing the square:

  1. $3x^2 + 3y^2 - 6x + 12y - 5 = 0$
  2. $2x^2 - 5y^2 + 4x + 10y - 7 = 0$
  3. $4x^2 + 9y^2 - 16x + 54y + 61 = 0$
  4. $x^2 - 6x + 8y + 1 = 0$

Solution.

  1. $A = 3$, $C = 3$, $A = C$ → Circle
  2. $A = 2$, $C = -5$, opposite signs → Hyperbola
  3. $A = 4$, $C = 9$, same sign and $A \neq C$ → Ellipse
  4. $A = 1$, $C = 0$ → Parabola

Example 8.5.2 — Using the General Discriminant

Classify the conic $x^2 + 4xy + 4y^2 + 5x - 3 = 0$.

Solution. Here $A = 1$, $B = 4$, $C = 4$.

$$B^2 - 4AC = 16 - 16 = 0$$

Since the discriminant equals zero, this is a parabola (possibly degenerate).

8.6 Rotation of Axes (Advanced)

When the general second-degree equation contains an $xy$-term ($B \neq 0$), the conic is rotated relative to the coordinate axes. To eliminate the $xy$-term, we introduce a rotated coordinate system $(x', y')$ using the substitution:

$$x = x'\cos\theta - y'\sin\theta, \quad y = x'\sin\theta + y'\cos\theta$$

where the rotation angle $\theta$ satisfies:

$$\cot 2\theta = \frac{A - C}{B}$$

Rotation of Axes Theorem

For any second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ with $B \neq 0$, there exists an angle $\theta$ such that in the rotated coordinate system $(x', y')$, the equation has no $x'y'$-term. The angle is determined by $\cot 2\theta = \dfrac{A - C}{B}$, where $0 < 2\theta < \pi$.

After performing the rotation, the equation takes the form $A'x'^2 + C'y'^2 + D'x' + E'y' + F' = 0$, which can then be analyzed using the standard techniques from Sections 8.2–8.4.

Example 8.6.1 — Eliminating the $xy$-term

Identify the conic $xy = 4$ by rotating axes to eliminate the $xy$-term.

Solution. Write the equation as $0 \cdot x^2 + 1 \cdot xy + 0 \cdot y^2 - 4 = 0$, so $A = 0$, $B = 1$, $C = 0$.

Find $\theta$: $\cot 2\theta = \dfrac{A - C}{B} = \dfrac{0 - 0}{1} = 0$, so $2\theta = 90°$ and $\theta = 45°$.

Substituting $x = \dfrac{x' - y'}{\sqrt{2}}$ and $y = \dfrac{x' + y'}{\sqrt{2}}$:

$$\frac{(x' - y')(x' + y')}{2} = 4 \implies \frac{x'^2 - y'^2}{2} = 4 \implies \frac{x'^2}{8} - \frac{y'^2}{8} = 1$$

This is a hyperbola centered at the origin in the rotated system, with $a = b = 2\sqrt{2}$.

Exam Tip: Rotation of axes problems are relatively rare on standardized exams. The most important skill is using the discriminant $B^2 - 4AC$ to classify the conic without performing the full rotation.

8.7 Applications of Conic Sections

Conic sections arise naturally in physics, engineering, astronomy, and architecture. Here are several notable applications:

Satellite Orbits (Ellipses)

By Kepler's First Law, every planet orbits the Sun in an ellipse with the Sun at one focus. The same applies to satellites orbiting Earth. If a satellite has a perigee (closest distance) of $r_1$ and an apogee (farthest distance) of $r_2$, then the semi-major axis is $a = \dfrac{r_1 + r_2}{2}$ and the distance from the center to the focus is $c = \dfrac{r_2 - r_1}{2}$.

Example 8.7.1 — Satellite Orbit

A satellite orbits Earth in an elliptical path with a perigee of 200 km and an apogee of 800 km above Earth's surface. If Earth's radius is approximately 6371 km, find the equation of the orbit (center at the center of the ellipse) and the eccentricity.

Solution. The closest distance from Earth's center: $r_1 = 6371 + 200 = 6571$ km. The farthest: $r_2 = 6371 + 800 = 7171$ km.

The equation (center at the center of the ellipse): $\dfrac{x^2}{6871^2} + \dfrac{y^2}{6858.6^2} = 1$.

Eccentricity: $e = \dfrac{c}{a} = \dfrac{300}{6871} \approx 0.0437$. This is very close to circular.

Reflective Properties (Parabolas and Ellipses)

A parabolic reflector focuses incoming parallel rays at the focus. This principle is used in satellite dishes, car headlights, and reflecting telescopes. An elliptical reflector reflects any signal emanating from one focus so that it passes through the other focus. This is exploited in lithotripsy (breaking kidney stones) and whispering galleries.

Example 8.7.2 — Parabolic Dish

A satellite dish is in the shape of a paraboloid. The dish is 2 meters wide and 0.5 meters deep. Where should the receiver be placed?

Solution. Set up coordinates with the vertex at the origin and the parabola opening upward: $x^2 = 4py$. The edge of the dish is at the point $(1, 0.5)$ (half the width, full depth).

Substituting: $1^2 = 4p(0.5) \implies 1 = 2p \implies p = 0.5$.

The receiver (focus) should be placed $0.5$ meters above the vertex of the dish.

LORAN Navigation (Hyperbolas)

Long Range Navigation (LORAN) uses the time difference of radio signals from two fixed transmitters to locate a ship or aircraft. The locus of points with a constant time difference from two transmitters forms one branch of a hyperbola. By using multiple pairs of transmitters, the position is determined as the intersection of hyperbolas.

8.8 Interactive Conic Explorer

Use the interactive graphs below to explore how the parameters $a$ and $b$ affect the shape of an ellipse and a hyperbola. Adjust the sliders and observe how the foci, vertices, and asymptotes change in real time.

Ellipse Explorer

Explore: Adjust $a$ and $b$ to see how the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ changes shape. The green points mark the foci.

In the ellipse explorer above, notice these behaviors:

Hyperbola Explorer

Explore: Adjust $a$ and $b$ to see how the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ changes shape. The dashed lines show the asymptotes.

In the hyperbola explorer above, observe:

8.9 Practice Problems

Test your understanding with these 12 problems covering all topics from this chapter. Click "Show Solution" to check your work.

Problem 1 — Parabola

Find the focus, directrix, and vertex of the parabola $y^2 = 20x$.

Show Solution

This is in the form $y^2 = 4px$. So $4p = 20$, giving $p = 5$.

  • Vertex: $(0, 0)$
  • Focus: $(5, 0)$
  • Directrix: $x = -5$

The parabola opens to the right.

Problem 2 — Parabola (Completing the Square)

Write $(x - 4)^2 = -8(y + 1)$ in general form, and find the focus and directrix.

Show Solution

We have $h = 4$, $k = -1$, and $4p = -8$, so $p = -2$.

  • Focus: $(4, -1 + (-2)) = (4, -3)$
  • Directrix: $y = -1 - (-2) = 1$

General form: $x^2 - 8x + 8y + 24 = 0$.

Problem 3 — Ellipse

Find the foci, vertices, and eccentricity of $\dfrac{x^2}{36} + \dfrac{y^2}{4} = 1$.

Show Solution

$a^2 = 36$, $b^2 = 4$, so $a = 6$, $b = 2$. Major axis is horizontal.

  • Vertices: $(\pm 6, 0)$
  • Foci: $c = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2}$, so foci at $(\pm 4\sqrt{2}, 0)$
  • Eccentricity: $e = \dfrac{4\sqrt{2}}{6} = \dfrac{2\sqrt{2}}{3} \approx 0.943$
Problem 4 — Ellipse (Equation from Conditions)

Write the equation of the ellipse with center at the origin, a vertex at $(0, 7)$, and a focus at $(0, 2\sqrt{6})$.

Show Solution

The vertex is on the $y$-axis, so the major axis is vertical. We have $a = 7$ and $c = 2\sqrt{6}$.

$b^2 = a^2 - c^2 = 49 - 24 = 25$, so $b = 5$.

$$\frac{x^2}{25} + \frac{y^2}{49} = 1$$
Problem 5 — Hyperbola

Find the vertices, foci, and asymptotes of $\dfrac{y^2}{25} - \dfrac{x^2}{144} = 1$.

Show Solution

The positive term is $y^2$, so the transverse axis is vertical. $a^2 = 25$, $b^2 = 144$, so $a = 5$, $b = 12$.

  • Vertices: $(0, \pm 5)$
  • Foci: $c = \sqrt{25 + 144} = \sqrt{169} = 13$, so foci at $(0, \pm 13)$
  • Asymptotes: $y = \pm \dfrac{5}{12}x$
Problem 6 — Hyperbola (Completing the Square)

Rewrite $9x^2 - 4y^2 - 36x - 8y - 4 = 0$ in standard form and find the center, vertices, and asymptotes.

Show Solution

Group and complete the square:

$$9(x^2 - 4x + 4) - 4(y^2 + 2y + 1) = 4 + 36 - 4 = 36$$ $$9(x - 2)^2 - 4(y + 1)^2 = 36$$ $$\frac{(x - 2)^2}{4} - \frac{(y + 1)^2}{9} = 1$$
  • Center: $(2, -1)$
  • $a = 2$, $b = 3$
  • Vertices: $(2 \pm 2, -1) = (4, -1)$ and $(0, -1)$
  • Asymptotes: $y + 1 = \pm \dfrac{3}{2}(x - 2)$
Problem 7 — Classifying Conics

Classify each conic: (a) $5x^2 + 5y^2 - 20x + 30y = 0$   (b) $3x^2 + 7y^2 - 12 = 0$   (c) $x^2 - 4y^2 + 6x = 0$

Show Solution

(a) $A = 5$, $C = 5$; $A = C$ → Circle

(b) $A = 3$, $C = 7$; same sign, $A \neq C$ → Ellipse

(c) $A = 1$, $C = -4$; opposite signs → Hyperbola

Problem 8 — General Discriminant

Classify $2x^2 + 6xy + 5y^2 - 10 = 0$ using the discriminant.

Show Solution

$A = 2$, $B = 6$, $C = 5$.

$$B^2 - 4AC = 36 - 40 = -4 < 0$$

Since the discriminant is negative, this is an ellipse.

Problem 9 — Rotation of Axes

Find the angle of rotation needed to eliminate the $xy$-term in $x^2 + 2\sqrt{3}\,xy + 3y^2 + 1 = 0$, and classify the resulting conic.

Show Solution

$A = 1$, $B = 2\sqrt{3}$, $C = 3$.

$\cot 2\theta = \dfrac{A - C}{B} = \dfrac{1 - 3}{2\sqrt{3}} = \dfrac{-2}{2\sqrt{3}} = \dfrac{-1}{\sqrt{3}}$, so $2\theta = 120°$ and $\theta = 60°$.

Discriminant: $B^2 - 4AC = 12 - 12 = 0$, so this is a parabola.

Problem 10 — Application (Satellite Orbit)

A satellite orbits Earth in an elliptical path. The perigee is 400 km and the apogee is 1200 km above Earth's surface (Earth's radius $\approx 6371$ km). Find the eccentricity of the orbit.

Show Solution

$r_1 = 6371 + 400 = 6771$ km, $r_2 = 6371 + 1200 = 7571$ km.

$a = \dfrac{r_1 + r_2}{2} = \dfrac{6771 + 7571}{2} = 7171$ km.

$c = \dfrac{r_2 - r_1}{2} = \dfrac{7571 - 6771}{2} = 400$ km.

$e = \dfrac{c}{a} = \dfrac{400}{7171} \approx 0.0558$.

Problem 11 — Application (Parabolic Reflector)

A flashlight reflector is parabolic with a diameter of 6 cm and a depth of 2 cm. How far from the vertex should the bulb be placed?

Show Solution

Model as $x^2 = 4py$ with vertex at the origin. The edge of the reflector is at $(3, 2)$.

$9 = 4p(2) = 8p \implies p = \dfrac{9}{8} = 1.125$ cm.

The bulb should be placed $1.125$ cm from the vertex.

Problem 12 — Mixed (Writing Equations)

Write the equation of the conic with foci at $(\pm 5, 0)$ and the sum of distances from any point on the curve to the foci equal to $14$.

Show Solution

The condition describes an ellipse (constant sum of distances). We have $c = 5$ and $2a = 14$, so $a = 7$.

$b^2 = a^2 - c^2 = 49 - 25 = 24$.

$$\frac{x^2}{49} + \frac{y^2}{24} = 1$$
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