Conic Sections — AP® Precalculus Lesson

Part 1 · Hook

Why this matters

Slice a cone with a flat cut and the edge of the cut is a curve. Tilt the cut and the curve changes: a circle, a stretched oval, a U-shape, or a pair of open branches. These conic sections show up as satellite dishes, planet orbits and the shadow a lampshade throws on a wall.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the vertex of $y = 2(x - 3)^2 + 1$ ?

$(3, 1)$
$(-3, 1)$
$(1, 3)$
$(3, 2)$

Show the answer

$y = a(x - h)^2 + k$ has vertex $(h, k) = (3, 1)$ .

Correct: $(3, 1)$ :
$(-3, 1)$ :
$(1, 3)$ :
$(3, 2)$ :

2. Which points of $x^2 + y^2 = 9$ have $y = 0$ ?

$(3, 0)$ and $(-3, 0)$
$(9, 0)$ and $(-9, 0)$
$(0, 3)$ and $(0, -3)$
$(3, 0)$ , the one point on the right

Show the answer

$x^2 = 9$ , so $x = 3$ or $x = -3$ .

Correct: $(3, 0)$ and $(-3, 0)$ :
$(9, 0)$ and $(-9, 0)$ :
$(0, 3)$ and $(0, -3)$ :
$(3, 0)$ , the one point on the right:

Part 4 · See it

See it first

Three graphs. Left: the parabola y = 0.5 x squared with its vertex at the origin. Middle: the ellipse x squared over 9 plus y squared over 4 = 1, centered at the origin, with vertices (3, 0) and (-3, 0) and the ends of the short axis at (0, 2) and (0, -2). Right: the hyperbola x squared minus y squared over 2.25 = 1, with vertices (1, 0) and (-1, 0), two branches opening left and right, and dashed asymptotes y = 1.5x and y = -1.5x. — A parabola, an ellipse and a hyperbola. The ellipse reaches $3$ across and $2$ up from its center; the hyperbola has two branches that approach the dashed asymptotes. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Only one variable is squared, as in $y = a(x - h)^2 + k$ the curve is a parabola with vertex $(h, k)$
Both squares are added and set equal to $1$ : $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2}$ $= 1$ an ellipse with center $(h, k)$ , reaching $a$ units across and $b$ units up and down
The squares are subtracted: $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2}$ $= 1$ a hyperbola with center $(h, k)$ , opening left and right, with asymptotes of slope $\pm\frac{b}{a}$
The equation is not in standard formcomplete the square in each variable, then read the conic off the signs

Part 6 · Key ideas

Key ideas

Parabola: $y - k = a(x - h)^2$ opens up or down; $x - h = a(y - k)^2$ opens right ( $a > 0$ ) or left.
Ellipse: $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2}$ $= 1$ , center $(h, k)$ ; vertices $a$ units across and $b$ units up and down. $a = b$ gives a circle.
Hyperbola: $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2}$ $= 1$ opens left and right, vertices $(h \pm a, k)$ , asymptotes $y - k = \pm\frac{b}{a}(x - h)$ . With $y$ first it opens up and down.
Both squares with the same sign: ellipse. Opposite signs: hyperbola. One square: parabola.

Part 7 · Misconception

A common mistake

The wrong idea: In $\frac{(x - 1)^2}{16} + \frac{(y + 2)^2}{4}$ $= 1$ the ellipse reaches 16 units left and right of its center.

What actually happens: The denominators are squares. $a^2 = 16$ , so it reaches $a = 4$ units left and right, and $b^2 = 4$ gives $2$ units up and down.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

An ellipse

1. What is the center of the ellipse?

$(1, -2)$
$(-1, 2)$
$(1, 2)$
$(-2, 1)$

Show the answer

The ellipse runs from $x = -3$ to $x = 5$ and from $y = -4$ to $y = 0$ . The middle of each range is $x = 1$ and $y = -2$ .

Correct: $(1, -2)$ : Midpoints of $[-3, 5]$ and $[-4, 0]$ .
$(-1, 2)$ : Both signs are flipped; the ellipse is below the $x$ -axis.
$(1, 2)$ : The center is below the axis, at $y = -2$ .
$(-2, 1)$ : The coordinates are swapped.

2. Which equation does the ellipse have?

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

Show the answer

Center $(1, -2)$ ; it reaches $4$ units across, so $a^2 = 16$ , and $2$ units up and down, so $b^2 = 4$ .

Correct: $\frac{(x - 1)^2}{16} + \frac{(y + 2)^2}{4}$ $= 1$ : Center $(1, -2)$ , $a = 4$ , $b = 2$ .
$\frac{(x - 1)^2}{4} + \frac{(y + 2)^2}{16}$ $= 1$ : This one is taller than wide; the graph is wider.
$\frac{(x + 1)^2}{16} + \frac{(y - 2)^2}{4}$ $= 1$ : $(x + 1)$ and $(y - 2)$ put the center at $(-1, 2)$ .
$\frac{(x - 1)^2}{4} + \frac{(y + 2)^2}{2}$ $= 1$ : The denominators are $a^2$ and $b^2$ , not $a$ and $b$ .

3. Which points are the vertices at the ends of the long axis?

$(5, -2)$ and $(-3, -2)$
$(1, 0)$ and $(1, -4)$
$(17, -2)$ and $(-15, -2)$
$(5, 2)$ and $(-3, 2)$

Show the answer

The long axis is horizontal: $(1 \pm 4, -2)$ , which is $(5, -2)$ and $(-3, -2)$ .

Correct: $(5, -2)$ and $(-3, -2)$ : $4$ units left and right of $(1, -2)$ .
$(1, 0)$ and $(1, -4)$ : These are the ends of the short axis, $2$ units up and down.
$(17, -2)$ and $(-15, -2)$ : Move $a = 4$ units, not $a^2 = 16$ .
$(5, 2)$ and $(-3, 2)$ : The vertices share the center's $y$ -coordinate, $-2$ .

4. What is the length of the long axis of the ellipse?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

From $(-3, -2)$ to $(5, -2)$ is $5 - (-3) = 8$ , which is $2a = 2(4)$ .

Answer: 8

5. Describe the graph of $y^2 - 4x^2 = 16$ .

A hyperbola opening up and down
A hyperbola opening left and right
An ellipse centered at the origin
A parabola opening to the right

Show the answer

Opposite signs: a hyperbola. Divide by $16$ : $\frac{y^2}{16} - \frac{x^2}{4} = 1$ . The $y$ term is positive, so it opens up and down, with vertices $(0, \pm 4)$ .

Correct: A hyperbola opening up and down: $y$ term positive: up and down.
A hyperbola opening left and right: Left and right needs the $x$ term positive.
An ellipse centered at the origin: An ellipse needs the same sign on both squares.
A parabola opening to the right: Both variables are squared, so it is not a parabola.

6. Rewrite $x^2 + y^2 - 6x + 4y - 3$ $= 0$ by completing the square. What is the graph?

A circle with center $(3, -2)$ and radius $4$
A circle with center $(-3, 2)$ and radius $4$
A circle with center $(3, -2)$ and radius $16$
A circle with center $(3, -2)$ and radius $\sqrt{3}$

Show the answer

$(x^2 - 6x + 9)$ $+ (y^2 + 4y + 4)$ $= 3$ $+ 9$ $+ 4$ , so $(x - 3)^2 + (y + 2)^2 = 16$ : center $(3, -2)$ , radius $4$ .

Correct: A circle with center $(3, -2)$ and radius $4$ : $(x - 3)^2 + (y + 2)^2 = 16$ .
A circle with center $(-3, 2)$ and radius $4$ : $(x - 3)^2$ means the center has $x = 3$ .
A circle with center $(3, -2)$ and radius $16$ : $16$ is the radius squared.
A circle with center $(3, -2)$ and radius $\sqrt{3}$ : Add the $9$ and $4$ to the right side too: $3 + 9 + 4 = 16$ .

Part 9 · Summary

Summary

Conic sections are the parabola, the ellipse and the hyperbola, all implicitly defined curves. In standard form you read off the center or vertex, the distances

a

and

b

from the denominators' square roots, and for a hyperbola the asymptotes

y - k = \pm\frac{b}{a}(x - h)

. The signs of the squared terms tell which conic you have.

Part 10 · Up next

What comes next

Part 11 · Connections