# Ellipse

An **ellipse** is a conic section, that resembles an oval, but is formally characterized by the following property: there exist two points $F_1$ and $F_2$ inside the ellipse (called focal points) such that for every point $P$ on the ellipse, the quantity $PF_1 + PF_2$ is constant $($where $PF_i$ denotes the distance from $P$ to $F_i).$ Ellipses are essentially circles that have been stretched, generally (not necessarily) along one axis. They are important objects in coordinate geometry, Euclidean geometry and number theory.

## Equation and Terminology

Suppose two points $F_1$ and $F_2$ are given, and one wishes to determine the locus of points $P$ such that $PF_1 + PF_2$ is some constant $k=2a$. The points $F_1$ and $F_2$ are called the *foci* of the ellipse (singular: *focus*). To simplify the calculations, one assumes $F_1 = (-c,0)$ and $F_2 = (c,0)$. Given the equation when $F_1$ and $F_2$ are of this form, one may retrieve the more general equation by rotating, dilating, and translating accordingly.

The condition imposed is precisely

$\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2 + y^2} = 2a.$

Isolating the leftmost radical and squaring both sides gives

$(x+c)^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2 + y^2} + (x-c)^2 + y^2.$

Again, isolating the remaining radical and simplifying yields

$\sqrt{(x-c)^2 + y^2} = a - \frac{c}{a} x .$

Finally, squaring both sides and rearranging gives

$\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1.$

Setting $b= \sqrt{a^2 - c^2}$, the equation is simply

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$

The

equation of an ellipsecentered at the origin $(0,0)$ is$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$

where $a$ and $b$ are real numbers. More generally, an ellipse centered at $(x', y') \in \mathbb{R}^2$ will have an equation of the form

$\frac{(x-x')^2}{a^2} + \frac{(y-y')^2}{b^2} = 1.$

The axis of the ellipse through $F_1$ and $F_2$ is called the *major axis* of the ellipse, and the axis perpendicular to the major axis is the *minor axis*. In the above notation, the length of the major axis is $2a$, since the ellipse meets the $x$-axis at precisely $(a,0)$ and $(-a,0)$. An analogous observation shows the length of the minor axis is $2b$. The parameter $c$ is called the *focal distance* of the ellipse and represents the distance from a focus to the center of the ellipse.

The *eccentricity* of the ellipse is defined as $\varepsilon= \frac ca$. This can be thought of as measuring how much the ellipse deviates from being a circle; the ellipse is a circle precisely when $\varepsilon = 0$, and otherwise one has $\varepsilon < 1$.

A tunnel opening is shaped like a half ellipse. The tunnel is 80 units wide and 25 units tall. Find the equation of the ellipse assuming it is centered at the origin.

The length of the major axis (which is on the $x$-axis since it's width) is 40 units, and the length of the minor axis (which is on the $y$-axis since it's height) is 25 units.

Since the ellipse (or half ellipse) is centered at the origin, the equation is

$\dfrac{x^{2}}{40^2 } + \dfrac{y^2}{25^2} = 1.\ _\square$

Triangle $ABC$ has coodinates $A= (-4, 0)$, $B= (4 , 0)$, and $C= (0 , 3)$.

Let $P$ be the point in the first quadrant such that $\triangle ABP$ has half the area of $\triangle ABC$ but both triangles have the same perimeter.

What is the length of $CP?$ If your solution is in a form of $\sqrt{d}$, submit $d$ as the answer.

What is the equation of this ellipse?

## Area Formula

Intuitively, an ellipse with major axis of length $2a$ and minor axis of length $2b$ is simply a circle of radius $a$ that has been squished/stretched along the $y$-axis by a factor of $\frac ba$. Accordingly, the area enclosed by this ellipse should be $\frac{b}{a} \cdot \pi a^2 = \pi a b$.

Although this is not a rigorous proof, the intuition is not difficult to turn into a precise argument. The upper half of this ellipse has equation

$y = b\sqrt{1 - \frac{x^2}{a^2}},$

so the area of the ellipse is

$A = 2b \int_{-a}^{a} \sqrt{1 - \frac{x^2}{a^2}} \, dx = 2\frac{b}{a} \int_{-a}^{a} \sqrt{a^2 - x^2} \, dx.$

But $\int_{-a}^{a} \sqrt{a^2 - x^2} \, dx$ is just half the area of the circle with equation $x^2 + y^2 = a^2$, which equals $\frac{1}{2} \pi a^2$. Thus,

$A = 2\frac{b}{a} \cdot \frac{1}{2} \pi a^2 = \pi a b.$

## Examples and Problems

If an ellipse's area is the same as the area of a circle with radius 4, what is the product of the ellipse's major and minor axes?

First, we would like to find the area of the circle with radius 4. Using the area formula of a circle, we get

$\pi r^{2} = \pi \times 4^{2} = 16\pi.$

Now that we know the area formula of an ellipse is $\pi ab$, we get that $ab=16$. Note that $a$ and $b$ are the major and minor radii of the ellipse, and what we actually want are the major and minor axes, which are $2a$ and $2b$.

Our last step is to find the product of $2a$ and $2b:$

$2a \times 2b = 4ab =4 \times 16 = 64.\ _\square$

## Applications

In astronomy, Kepler's laws state that the orbit of a planet around the sun traces an ellipse, one of whose foci is the sun itself. Furthermore, information about this ellipse can quantify the orbital period of the planet (how much time it takes for the planet to go once around the sun).

If $p$ is the orbital period and the ellipse corresponding to this orbit has a major axis of length $2a$, then $p^2 \propto a^3$, where $\propto$ indicates direct proportionality.

The planet Xabros is orbiting the sun in an elliptical path with a major axis of length $2a$. Doofenschmirz has a machine that can scale the length of the ellipse's major axis by a factor of $k$: that is, after using the machine, Xabros will be in a path with a major axis of length $2ka$.

If Doofenschmirz's evil plan is to double the orbital period of Xabros, what is $k?$