Chapter 10: Conic Sections
Chapter Contents
Conic sections are the curves formed when a plane intersects a double cone. Depending on the angle of the cut, the resulting shape is a circle, ellipse, parabola, or hyperbola. These four curves appear throughout science and engineering: satellite orbits follow ellipses, headlight reflectors use parabolas, and GPS navigation relies on hyperbolic positioning. In this chapter you will learn to write and manipulate the equations of all four conics, graph them by hand, and solve systems that mix conics with lines or with each other.
10.1 Circles
A circle is the set of all points in a plane that are equidistant from a fixed point called the center. That constant distance is the radius.
Definition: Standard Form of a Circle
A circle with center $(h, k)$ and radius $r$ has the equation:
$$(x - h)^2 + (y - k)^2 = r^2$$
When the center is the origin, this simplifies to $x^2 + y^2 = r^2$.
General Form
Expanding the standard form produces the general form of a circle:
$$x^2 + y^2 + Dx + Ey + F = 0$$Notice that in the general form the coefficients of $x^2$ and $y^2$ are both 1 (or, more generally, equal to each other). This distinguishes a circle from the other conics. To recover the center and radius from the general form, you complete the square for both $x$ and $y$.
Completing the Square
Completing the square is the key algebraic technique for converting between general and standard form. The idea: for an expression $x^2 + bx$, add and subtract $(b/2)^2$ so the expression becomes a perfect square trinomial.
Example 1: Converting General Form to Standard Form
Find the center and radius of the circle $x^2 + y^2 - 6x + 4y - 12 = 0$.
Step 1. Group $x$-terms and $y$-terms, move the constant to the right:
$(x^2 - 6x) + (y^2 + 4y) = 12$
Step 2. Complete the square for each group. For $x$: half of $-6$ is $-3$, squared is $9$. For $y$: half of $4$ is $2$, squared is $4$. Add these to both sides:
$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$
Step 3. Factor and simplify:
$(x - 3)^2 + (y + 2)^2 = 25$
Result: Center $(3, -2)$, radius $r = 5$.
Example 2: Writing the Equation from Given Information
Write the equation of the circle with center $(-1, 5)$ and passing through the point $(2, 9)$.
Step 1. Find the radius using the distance formula:
$$r = \sqrt{(2 - (-1))^2 + (9 - 5)^2} = \sqrt{9 + 16} = 5$$
Step 2. Substitute into standard form:
$$(x + 1)^2 + (y - 5)^2 = 25$$
10.2 Parabolas (Conic Form)
You have worked with parabolas since Algebra 1, where $y = ax^2 + bx + c$ described a U-shaped curve. In the conic-section framework, a parabola is defined geometrically: it is the set of all points equidistant from a fixed point (the focus) and a fixed line (the directrix).
Definition: Parabola (Conic Definition)
A parabola is the locus of points $(x, y)$ such that the distance to the focus equals the distance to the directrix. The parameter $p$ is the directed distance from the vertex to the focus.
Vertical Parabolas
A parabola with vertex $(h, k)$ that opens upward (when $p > 0$) or downward (when $p < 0$) has the equation:
$$y = \frac{1}{4p}(x - h)^2 + k$$- Focus: $(h,\, k + p)$
- Directrix: $y = k - p$
- Axis of symmetry: $x = h$
Horizontal Parabolas
A parabola with vertex $(h, k)$ that opens right (when $p > 0$) or left (when $p < 0$) has the equation:
$$x = \frac{1}{4p}(y - k)^2 + h$$- Focus: $(h + p,\, k)$
- Directrix: $x = h - p$
- Axis of symmetry: $y = k$
Key Relationship: Direction and Sign of p
The sign of $p$ determines which way the parabola opens. For a vertical parabola, $p > 0$ opens up and $p < 0$ opens down. For a horizontal parabola, $p > 0$ opens right and $p < 0$ opens left. The latus rectum (the chord through the focus perpendicular to the axis) has length $|4p|$.
Example 3: Finding Focus and Directrix
Find the vertex, focus, and directrix of the parabola $y = \frac{1}{8}(x - 2)^2 - 3$.
Step 1. Identify the vertex: $(h, k) = (2, -3)$.
Step 2. Compare $\frac{1}{8}$ with $\frac{1}{4p}$: so $4p = 8$ and $p = 2$.
Step 3. Since $p > 0$, the parabola opens upward.
Focus: $(2,\, -3 + 2) = (2, -1)$.
Directrix: $y = -3 - 2 = -5$.
Example 4: Writing the Equation of a Horizontal Parabola
Write the equation of the parabola with vertex $(1, 3)$ and focus $(-2, 3)$.
Step 1. The focus and vertex share the same $y$-coordinate, so this is a horizontal parabola.
Step 2. $p = -2 - 1 = -3$ (focus is left of vertex, so $p < 0$).
Step 3. Use the horizontal form:
$$x = \frac{1}{4(-3)}(y - 3)^2 + 1 = -\frac{1}{12}(y - 3)^2 + 1$$
The parabola opens to the left, and the directrix is $x = 1 - (-3) = 4$.
10.3 Ellipses
An ellipse is the set of all points for which the sum of the distances to two fixed points (the foci) is constant. When both foci coincide the ellipse degenerates into a circle, so a circle is a special case of an ellipse.
Definition: Standard Form of an Ellipse
An ellipse centered at $(h, k)$ with semi-major axis $a$ and semi-minor axis $b$ (where $a > b > 0$) has one of two standard forms:
Horizontal major axis:
$$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$
Vertical major axis:
$$\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$$
The larger denominator is always $a^2$, and the major axis runs along whichever variable has $a^2$ beneath it.
Key Features of an Ellipse
- Center: $(h, k)$
- Vertices: endpoints of the major axis, a distance $a$ from the center
- Co-vertices: endpoints of the minor axis, a distance $b$ from the center
- Foci: located on the major axis, a distance $c$ from the center, where $c^2 = a^2 - b^2$
- Eccentricity: $e = \dfrac{c}{a}$, a measure of how elongated the ellipse is ($0 \le e < 1$)
Theorem: Relationship Between a, b, and c for Ellipses
For any ellipse, the semi-major axis $a$, semi-minor axis $b$, and focal distance $c$ satisfy:
$$c^2 = a^2 - b^2$$
Since $a > b$, the value $c$ is always real and positive. The closer $c$ is to $a$, the more elongated the ellipse; when $c = 0$, the ellipse is a circle.
Example 5: Graphing an Ellipse
Identify the center, vertices, foci, and eccentricity of $\dfrac{(x + 1)^2}{25} + \dfrac{(y - 2)^2}{9} = 1$.
Center: $(h, k) = (-1, 2)$.
Since $a^2 = 25 > b^2 = 9$, the major axis is horizontal: $a = 5$, $b = 3$.
Vertices: $(-1 \pm 5,\, 2)$, i.e., $(4, 2)$ and $(-6, 2)$.
Co-vertices: $(-1,\, 2 \pm 3)$, i.e., $(-1, 5)$ and $(-1, -1)$.
Foci: $c = \sqrt{25 - 9} = 4$, so the foci are $(-1 \pm 4,\, 2)$, i.e., $(3, 2)$ and $(-5, 2)$.
Eccentricity: $e = \dfrac{4}{5} = 0.8$. This ellipse is fairly elongated.
Example 6: Converting to Standard Form
Write $4x^2 + 9y^2 - 16x + 54y + 61 = 0$ in standard form.
Step 1. Group and factor: $4(x^2 - 4x) + 9(y^2 + 6y) = -61$.
Step 2. Complete the square: $4(x^2 - 4x + 4) + 9(y^2 + 6y + 9) = -61 + 16 + 81$.
Step 3. Factor and simplify: $4(x - 2)^2 + 9(y + 3)^2 = 36$.
Step 4. Divide both sides by 36:
$$\frac{(x - 2)^2}{9} + \frac{(y + 3)^2}{4} = 1$$
Center $(2, -3)$, $a = 3$ (horizontal), $b = 2$, $c = \sqrt{5}$.
Interactive: Ellipse explorer with adjustable $a$, $b$, $h$, and $k$. Drag the sliders to see how the shape, foci, and vertices change.
10.4 Hyperbolas
A hyperbola is the set of all points for which the absolute difference of the distances to two fixed points (the foci) is constant. Unlike an ellipse, a hyperbola consists of two separate branches that open away from each other.
Definition: Standard Form of a Hyperbola
Horizontal transverse axis (opens left and right):
$$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$$
Vertical transverse axis (opens up and down):
$$\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$$
In both forms, $a^2$ is the denominator beneath the positive term.
Key Features of a Hyperbola
- Center: $(h, k)$
- Vertices: a distance $a$ from the center along the transverse axis
- Foci: a distance $c$ from the center along the transverse axis, where $c^2 = a^2 + b^2$
- Transverse axis: the segment connecting the two vertices (length $2a$)
- Conjugate axis: perpendicular to the transverse axis (length $2b$)
- Asymptotes (horizontal transverse axis): $y - k = \pm\dfrac{b}{a}(x - h)$
- Asymptotes (vertical transverse axis): $y - k = \pm\dfrac{a}{b}(x - h)$
Theorem: Relationship Between a, b, and c for Hyperbolas
For any hyperbola, the values $a$, $b$, and $c$ satisfy:
$$c^2 = a^2 + b^2$$
Note the plus sign, in contrast with the minus sign for ellipses. Because of this, $c > a$ always holds for a hyperbola, and the foci lie outside the vertices.
Example 7: Analyzing a Hyperbola
Find the center, vertices, foci, and asymptotes of $\dfrac{(x - 3)^2}{16} - \dfrac{(y + 1)^2}{9} = 1$.
Center: $(3, -1)$.
Since the $x$-term is positive: horizontal transverse axis, $a^2 = 16$, $b^2 = 9$, so $a = 4$, $b = 3$.
Vertices: $(3 \pm 4,\, -1)$, i.e., $(7, -1)$ and $(-1, -1)$.
Foci: $c = \sqrt{16 + 9} = 5$, so the foci are $(3 \pm 5,\, -1)$, i.e., $(8, -1)$ and $(-2, -1)$.
Asymptotes: $y + 1 = \pm\dfrac{3}{4}(x - 3)$, which simplify to $y = \dfrac{3}{4}x - \dfrac{13}{4}$ and $y = -\dfrac{3}{4}x + \dfrac{5}{4}$.
Example 8: Writing a Hyperbola Equation
Write the standard-form equation of the hyperbola with vertices $(0, \pm 5)$ and foci $(0, \pm 13)$.
Step 1. Center is the midpoint of the vertices: $(0, 0)$. The transverse axis is vertical.
Step 2. $a = 5$ and $c = 13$, so $b^2 = c^2 - a^2 = 169 - 25 = 144$ and $b = 12$.
Step 3. Vertical transverse axis form:
$$\frac{y^2}{25} - \frac{x^2}{144} = 1$$
Asymptotes: $y = \pm\dfrac{5}{12}x$.
Interactive: Hyperbola explorer with adjustable $a$ and $b$. The dashed lines show the asymptotes, and the green points mark the foci.
10.5 Identifying Conic Sections
Every conic section can be written in the general second-degree form:
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$In Algebra 2 most problems have $B = 0$ (no $xy$-term), which means the axes of the conic are parallel to the coordinate axes. When $B = 0$, the type of conic is determined by the coefficients $A$ and $C$.
Classification When B = 0
Given $Ax^2 + Cy^2 + Dx + Ey + F = 0$ (with $A$ and $C$ not both zero):
Circle: $A = C$ (and $A \neq 0$)
Ellipse: $A \neq C$ but $A$ and $C$ have the same sign
Hyperbola: $A$ and $C$ have opposite signs
Parabola: $A = 0$ or $C = 0$ (but not both)
The Discriminant Method
For the fully general equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, the discriminant $\Delta = B^2 - 4AC$ classifies the conic:
| Discriminant | Conic Type |
|---|---|
| $B^2 - 4AC < 0$ | Ellipse (or circle if $A = C$ and $B = 0$) |
| $B^2 - 4AC = 0$ | Parabola |
| $B^2 - 4AC > 0$ | Hyperbola |
Degenerate cases (a single point, a line, or two intersecting lines) can occur when the equation factors or has no real solutions, but these are rare in standard Algebra 2 problems.
Example 9: Identifying a Conic
Classify $3x^2 + 3y^2 - 12x + 18y - 9 = 0$.
Here $A = 3$, $B = 0$, $C = 3$. Since $A = C$ and $B = 0$, this is a circle.
To confirm, divide by 3: $x^2 + y^2 - 4x + 6y - 3 = 0$. Completing the square: $(x - 2)^2 + (y + 3)^2 = 16$. Circle with center $(2, -3)$ and radius 4.
Example 10: Using the Discriminant
Classify $2x^2 + 5xy - 3y^2 + x - 4y + 7 = 0$.
$A = 2$, $B = 5$, $C = -3$.
$\Delta = B^2 - 4AC = 25 - 4(2)(-3) = 25 + 24 = 49 > 0$.
Since $\Delta > 0$, this is a hyperbola.
10.6 Systems with Conics
A system involving conic sections can have 0, 1, 2, 3, or 4 real solutions, depending on how the curves intersect. The two most common types in Algebra 2 are:
- Line and conic: Substitute the linear equation into the conic to obtain a single-variable quadratic.
- Two conics: Use elimination (subtract one equation from the other) or substitution to reduce the system.
Line-Conic Systems
Example 11: Line and Circle
Solve the system: $\begin{cases} x^2 + y^2 = 25 \\ y = x + 1 \end{cases}$
Step 1. Substitute $y = x + 1$ into the circle equation:
$x^2 + (x + 1)^2 = 25$
$x^2 + x^2 + 2x + 1 = 25$
$2x^2 + 2x - 24 = 0$
$x^2 + x - 12 = 0$
Step 2. Factor: $(x + 4)(x - 3) = 0$, so $x = -4$ or $x = 3$.
Step 3. Find corresponding $y$-values: $y = -4 + 1 = -3$ and $y = 3 + 1 = 4$.
Solutions: $(-4, -3)$ and $(3, 4)$. The line intersects the circle at two points.
Conic-Conic Systems
Example 12: Ellipse and Hyperbola
Solve the system: $\begin{cases} x^2 + 4y^2 = 25 \\ x^2 - y^2 = 5 \end{cases}$
Step 1. Subtract the second equation from the first to eliminate $x^2$:
$(x^2 + 4y^2) - (x^2 - y^2) = 25 - 5$
$5y^2 = 20$
$y^2 = 4$, so $y = \pm 2$.
Step 2. Substitute back: $x^2 = 5 + y^2 = 5 + 4 = 9$, so $x = \pm 3$.
Solutions: $(3, 2)$, $(-3, 2)$, $(3, -2)$, $(-3, -2)$ — four intersection points.
Maximum Number of Intersections
A line can intersect a conic in at most 2 points. Two distinct conics can intersect in at most 4 points. These maxima follow from Bézout's theorem: two curves of degrees $m$ and $n$ intersect in at most $mn$ points (counting multiplicity).
Summary: Conic Sections at a Glance
| Conic | Standard Form | Key Relationship | Eccentricity |
|---|---|---|---|
| Circle | $(x-h)^2+(y-k)^2=r^2$ | All points equidistant from center | $e = 0$ |
| Parabola | $y = \frac{1}{4p}(x-h)^2+k$ | Dist. to focus = Dist. to directrix | $e = 1$ |
| Ellipse | $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ | $c^2 = a^2 - b^2$ | $0 < e < 1$ |
| Hyperbola | $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ | $c^2 = a^2 + b^2$ | $e > 1$ |
10.7 Practice Problems
Work through these problems, then expand the solution to check your work. Complete solutions are provided with all intermediate steps.
Problem 1
Find the center and radius of the circle $x^2 + y^2 + 10x - 8y + 16 = 0$.
Show Solution
Group terms: $(x^2 + 10x) + (y^2 - 8y) = -16$.
Complete the square: $(x^2 + 10x + 25) + (y^2 - 8y + 16) = -16 + 25 + 16$.
$(x + 5)^2 + (y - 4)^2 = 25$.
Center: $(-5, 4)$, Radius: $r = 5$.
Problem 2
Find the focus and directrix of $y = -\frac{1}{16}(x + 3)^2 + 5$.
Show Solution
Vertex: $(-3, 5)$. Compare $-\frac{1}{16}$ with $\frac{1}{4p}$: $4p = -16$, so $p = -4$.
Since $p < 0$, the parabola opens downward.
Focus: $(-3, 5 + (-4)) = (-3, 1)$.
Directrix: $y = 5 - (-4) = 9$.
Problem 3
Write the equation of the parabola with vertex at the origin and focus at $(0, 6)$.
Show Solution
Vertex $(0,0)$, focus $(0,6)$: vertical parabola opening upward, $p = 6$.
$$y = \frac{1}{4(6)}x^2 = \frac{1}{24}x^2$$
Problem 4
For the ellipse $\frac{(x-1)^2}{49} + \frac{(y+2)^2}{25} = 1$, find the center, vertices, foci, and eccentricity.
Show Solution
Center: $(1, -2)$. $a^2 = 49$, $b^2 = 25$, so $a = 7$, $b = 5$. Major axis is horizontal.
Vertices: $(1 \pm 7, -2)$, i.e., $(8, -2)$ and $(-6, -2)$.
Foci: $c = \sqrt{49 - 25} = \sqrt{24} = 2\sqrt{6}$. Foci at $(1 \pm 2\sqrt{6}, -2)$.
Eccentricity: $e = \frac{2\sqrt{6}}{7} \approx 0.700$.
Problem 5
Convert $9x^2 + 25y^2 - 36x + 50y - 164 = 0$ to standard form and identify the conic.
Show Solution
Group: $9(x^2 - 4x) + 25(y^2 + 2y) = 164$.
Complete the square: $9(x^2 - 4x + 4) + 25(y^2 + 2y + 1) = 164 + 36 + 25 = 225$.
$9(x - 2)^2 + 25(y + 1)^2 = 225$.
Divide by 225: $\frac{(x - 2)^2}{25} + \frac{(y + 1)^2}{9} = 1$.
This is an ellipse centered at $(2, -1)$ with $a = 5$ (horizontal) and $b = 3$.
Problem 6
Find the vertices, foci, and asymptotes of $\frac{y^2}{36} - \frac{x^2}{64} = 1$.
Show Solution
Vertical transverse axis (the $y$-term is positive). Center: $(0, 0)$.
$a^2 = 36$, $b^2 = 64$, so $a = 6$, $b = 8$.
Vertices: $(0, \pm 6)$.
Foci: $c = \sqrt{36 + 64} = 10$. Foci at $(0, \pm 10)$.
Asymptotes: $y = \pm\frac{a}{b}x = \pm\frac{6}{8}x = \pm\frac{3}{4}x$.
Problem 7
Classify the conic: $4x^2 - 9y^2 + 24x + 36y - 36 = 0$.
Show Solution
$A = 4$, $C = -9$. Since $A$ and $C$ have opposite signs, this is a hyperbola.
To verify, complete the square:
$4(x^2 + 6x) - 9(y^2 - 4y) = 36$.
$4(x^2 + 6x + 9) - 9(y^2 - 4y + 4) = 36 + 36 - 36 = 36$.
$4(x + 3)^2 - 9(y - 2)^2 = 36$.
$$\frac{(x + 3)^2}{9} - \frac{(y - 2)^2}{4} = 1$$
Hyperbola centered at $(-3, 2)$, $a = 3$, $b = 2$.
Problem 8
Use the discriminant to classify $5x^2 + 3xy + 2y^2 - x + y - 10 = 0$.
Show Solution
$A = 5$, $B = 3$, $C = 2$.
$\Delta = B^2 - 4AC = 9 - 4(5)(2) = 9 - 40 = -31 < 0$.
Since $\Delta < 0$ and $A \neq C$, this is an ellipse (rotated, due to the $xy$-term).
Problem 9
Solve the system: $\begin{cases} x^2 + y^2 = 20 \\ y = 2x \end{cases}$
Show Solution
Substitute $y = 2x$ into the circle equation:
$x^2 + (2x)^2 = 20 \implies 5x^2 = 20 \implies x^2 = 4 \implies x = \pm 2$.
When $x = 2$: $y = 4$. When $x = -2$: $y = -4$.
Solutions: $(2, 4)$ and $(-2, -4)$.
Problem 10
Solve the system: $\begin{cases} x^2 + y^2 = 34 \\ x^2 - y^2 = 16 \end{cases}$
Show Solution
Add the two equations: $2x^2 = 50$, so $x^2 = 25$ and $x = \pm 5$.
Subtract the second from the first: $2y^2 = 18$, so $y^2 = 9$ and $y = \pm 3$.
Solutions: $(5, 3)$, $(5, -3)$, $(-5, 3)$, $(-5, -3)$.
Check: $25 + 9 = 34$ and $25 - 9 = 16$. Both equations satisfied.
Problem 11
An ellipse has foci at $(\pm 3, 0)$ and passes through the point $(5, 0)$. Write its equation in standard form.
Show Solution
Center: $(0, 0)$ (midpoint of foci). Horizontal major axis. $c = 3$.
Since $(5, 0)$ is on the ellipse and lies on the major axis, it is a vertex: $a = 5$.
$b^2 = a^2 - c^2 = 25 - 9 = 16$.
$$\frac{x^2}{25} + \frac{y^2}{16} = 1$$
Problem 12
A hyperbola has vertices at $(2, 1)$ and $(2, 7)$ and one focus at $(2, -1)$. Write its equation and find the asymptotes.
Show Solution
Center: midpoint of vertices $= (2, 4)$. Vertical transverse axis.
$a =$ distance from center to vertex $= |7 - 4| = 3$.
$c =$ distance from center to focus $= |{-1} - 4| = 5$.
$b^2 = c^2 - a^2 = 25 - 9 = 16$, so $b = 4$.
$$\frac{(y - 4)^2}{9} - \frac{(x - 2)^2}{16} = 1$$
Asymptotes: $y - 4 = \pm\frac{3}{4}(x - 2)$, i.e., $y = \frac{3}{4}x + \frac{5}{2}$ and $y = -\frac{3}{4}x + \frac{11}{2}$.