Honors Pre-Calculus Unit 10 Review: Analytic Geometry

An ellipse is the set of all points whose distances to two fixed points, the foci, add up to the same constant. That constant equals $2a$, the length of the major axis.

The standard form is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ when the major axis is horizontal, with center $(h, k)$. If the major axis is vertical, $a^2$ moves under the $y$ term. Most books keep $a > b$, so the larger denominator is $a^2$, and its variable tells you which way the long axis runs.

The foci sit on the major axis, $c$ units from the center, where $c^2 = a^2 - b^2$. Eccentricity $e = \frac{c}{a}$ falls between 0 and 1. A value near 0 means nearly a circle, and a value near 1 means long and skinny.

Worked example. Graph $\frac{(x-2)^2}{25} + \frac{(y+1)^2}{9} = 1$. The center is $(2, -1)$, since $y+1$ means $k = -1$. The larger denominator, 25, is under $x$, so the major axis is horizontal with $a = 5$ and $b = 3$. The vertices are $(2 \pm 5, -1)$, which gives $(7, -1)$ and $(-3, -1)$. The co-vertices are $(2, 2)$ and $(2, -4)$. Then $c^2 = 25 - 9 = 16$, so $c = 4$ and the foci are $(6, -1)$ and $(-2, -1)$. The eccentricity is $\frac{4}{5} = 0.8$.

Planets orbit the Sun on ellipses with the Sun at one focus. Earth's orbit has $e \approx 0.017$, so it is almost a circle.

A hyperbola is the set of points where the absolute difference of the distances to two foci is constant, equal to $2a$. Changing the sum to a difference flips the shape into two separate branches.

The positive term tells you which way it opens. $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ opens left and right, and $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ opens up and down. Here $a^2$ always goes with the positive term, even when it is the smaller number.

For hyperbolas, $c^2 = a^2 + b^2$, so the foci sit farther from the center than the vertices. The asymptotes pass through the center. For a horizontal hyperbola they are $y - k = \pm \frac{b}{a}(x - h)$, and for a vertical one they are $y - k = \pm \frac{a}{b}(x - h)$. To sketch, draw a central box with length $2a$ along the transverse axis and length $2b$ along the perpendicular axis. Extend its diagonals as the asymptotes, then draw the branches through the vertices.

Worked example. For $\frac{(x-3)^2}{16} - \frac{(y+1)^2}{9} = 1$, the center is $(3, -1)$, $a = 4$, $b = 3$, and it opens horizontally. The vertices are $(7, -1)$ and $(-1, -1)$. Since $c = \sqrt{16 + 9} = 5$, the foci are $(8, -1)$ and $(-2, -1)$. The asymptotes are $y + 1 = \pm \frac{3}{4}(x - 3)$, and $e = \frac{5}{4} = 1.25$.

Location systems like LORAN used hyperbolas, because a fixed difference in signal arrival times from two stations puts a ship somewhere on one hyperbola.

A parabola is the set of points that are the same distance from a fixed point, the focus, and a fixed line, the directrix. The vertex sits exactly halfway between them.

In this unit, parabolas are written in a form that shows the focus. A parabola with vertex $(h, k)$ that opens up or down is $(x-h)^2 = 4p(y-k)$. One that opens right or left is $(y-k)^2 = 4p(x-h)$. The number $p$ is the directed distance from the vertex to the focus. If $p > 0$ it opens up or right, and if $p < 0$ it opens down or left. The directrix is $|p|$ units from the vertex on the opposite side.

For a vertical parabola, this matches the vertex form $y = a(x-h)^2 + k$ from earlier units, with $a = \frac{1}{4p}$. A small $|p|$ means a narrow parabola.

Worked example. Find the focus and directrix of $y^2 = -8x$. Only $y$ is squared, so it opens sideways. Matching $4p = -8$ gives $p = -2$, so it opens left. The vertex is $(0, 0)$, the focus is $(-2, 0)$, and the directrix is the line $x = 2$. Check with the point $(-2, 4)$, which satisfies $16 = -8(-2)$. It is 4 units from the focus and 4 units from the line $x = 2$.

The focus explains why satellite dishes and headlights are parabolic. Signals arriving parallel to the axis all reflect to the focus, and light from a bulb at the focus reflects out as a parallel beam.

When a nondegenerate conic's equation has an $xy$ term, its axis is tilted relative to the coordinate axes. In the general form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, the $xy$ term is present when $B \neq 0$. The equation can also describe a degenerate case, such as intersecting lines.

For a nondegenerate real conic, you can identify the type before rotating by using the discriminant $B^2 - 4AC$. A negative value means an ellipse or circle, zero means a parabola, and a positive value means a hyperbola. The discriminant stays the same under rotation, but it does not rule out degenerate cases or equations with no real points.

To remove the $xy$ term, rotate the axes by an angle $\theta$ with $\cot(2\theta) = \frac{A - C}{B}$. Then substitute $x = x'\cos\theta - y'\sin\theta$ and $y = x'\sin\theta + y'\cos\theta$ and simplify. The new equation in $x'$ and $y'$ has no cross term, so you can complete the square and read it like any standard conic.

Worked example. Classify and rotate $xy = 1$, which is $xy - 1 = 0$. Here $A = 0$, $B = 1$, and $C = 0$, so $B^2 - 4AC = 1 > 0$, a hyperbola. Since $A = C$, $\cot(2\theta) = 0$, so $2\theta = 90°$ and $\theta = 45°$. With $\cos 45° = \sin 45° = \frac{\sqrt{2}}{2}$, you get $x = \frac{x' - y'}{\sqrt{2}}$ and $y = \frac{x' + y'}{\sqrt{2}}$. Multiplying gives $xy = \frac{x'^2 - y'^2}{2} = 1$, or $\frac{x'^2}{2} - \frac{y'^2}{2} = 1$. That is a hyperbola with $a = \sqrt{2}$, opening along the line $y = x$.

With one focus at the pole (the origin) and the conic's axis along a coordinate axis, an ellipse, parabola, or hyperbola has the polar equation $r = \frac{ep}{1 \pm e\cos\theta}$ or $r = \frac{ep}{1 \pm e\sin\theta}$. Here $e > 0$ is the eccentricity and $p$ is the distance from the focus to the directrix. This use of $p$ differs from the directed vertex-to-focus distance in the parabola formulas.

This comes from the focus-directrix definition for ellipses, parabolas, and hyperbolas. A point is on the conic when its distance to the focus divided by its distance to the directrix equals $e$. So $0 < e < 1$ gives an ellipse, $e = 1$ a parabola, and $e > 1$ a hyperbola. A circle has $e = 0$ but is not produced by this ratio.

The trig function and the sign locate the directrix. With $+\cos\theta$ the directrix is $x = p$, and with $-\cos\theta$ it is $x = -p$. With $+\sin\theta$ it is $y = p$, and with $-\sin\theta$ it is $y = -p$. Before reading off $e$, make sure the constant in the denominator is 1.

Worked example. Identify $r = \frac{6}{3 + 2\cos\theta}$. Divide the top and bottom by 3 to get $r = \frac{2}{1 + \frac{2}{3}\cos\theta}$. So $e = \frac{2}{3}$, which makes it an ellipse. Since $ep = 2$, $p = 3$ and the directrix is $x = 3$. The vertices come from $\theta = 0$, which gives $r = \frac{6}{5} = 1.2$ and the point $(1.2, 0)$, and from $\theta = \pi$, which gives $r = 6$ and the point $(-6, 0)$. The major axis is $1.2 + 6 = 7.2$ units long, so $a = 3.6$.