The Ellipse

Try this

Drag $P$ once around the ellipse. The two pieces of string $∣ P F_{1} ∣$ and $∣ P F_{2} ∣$ each change. What about their sum?
Drag $c$ to $0$ : the two pins coincide. What do you draw? Now slowly increase $c$ . How do the eccentricity $e$ and the shape of the ellipse change together?
Drag $c$ until it equals $a$ , then beyond $a$ . Can the string still trace an ellipse?
Tick “Let $P$ leave the curve”. Drag $P$ inside the ellipse and outside it, and compare the sum with $2 a$ .

Pin the two ends of a string at $F_{1}, F_{2}$ , keep the string taut with the tip of a pencil and go once around: what you draw is an ellipse:

∣ P F_{1} ∣ + ∣ P F_{2} ∣ = 2 a (2 a > ∣ F_{1} F_{2} ∣ = 2 c)

$F_{1}, F_{2}$ are called the foci, and $2 c$ is the focal distance.

Predict first

A point $P$ on the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ is at distance $3$ from the left focus. How far is it from the right focus?

Choose an answer first; then you can see the result.

Why this equation

Take the line $F_{1} F_{2}$ as the $x$ -axis and its midpoint as the origin, and translate “the string length does not change” into coordinates. Squaring twice gets rid of the two square roots.

From the definition to the standard equation

Step 1 / 6

Set up coordinates
The foci are $F_{1} (- c, 0)$ and $F_{2} (c, 0)$ ; in the figure $c = 3$ and $a = 5$ . $P (x, y)$ is on the ellipse if and only if the two distances add up to $2 a$ .

$(x + c)^{2} + y^{2} + (x - c)^{2} + y^{2} = 2 a$
Move one root across, square once
Square a second time
Write b² = a² − c²
The standard equation
Eccentricity

If the foci are on the $y$ -axis, just swap $x$ and $y$ : $\frac{y ^{2}}{a ^{2}} + \frac{x ^{2}}{b ^{2}} = 1$ . How to tell: the major axis lies on the axis with the larger denominator.

An ellipse is a squashed circle

Multiply the $y$ -coordinate of every point of the circle $x^{2} + y^{2} = a^{2}$ by $\frac{b}{a}$ . The resulting point $(x, y^{'})$ satisfies $x^{2} + (\frac{a}{b} y^{'})^{2} = a^{2}$ , which is exactly $\frac{x ^{2}}{a ^{2}} + \frac{y ^{'2}}{b ^{2}} = 1$ . So the area of the ellipse is $π a^{2} \cdot \frac{b}{a} = π ab$ . As $e \to 0$ , $b \to a$ , the squashing ratio tends to $1$ , and the ellipse turns back into a circle.

If the string is too short

Break it

Counterexample lab: remove one condition

Conclusion: The points whose distances to two fixed points add up to $2 a$ form an ellipse

$a > c$ : drag $P$ once around and you draw a complete ellipse.

Practice

Exercise 1Basic

Find the focal distance $2 c$ of the ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1$ .

Exercise 2Basic

Find the eccentricity of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ (as a fraction).

Exercise 3Advanced

$F_{1}, F_{2}$ are the two foci of the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1$ , and $P$ is a point of the ellipse not on the $x$ -axis. Find the perimeter of $△ P F_{1} F_{2}$ .

Exercise 4Challenge

$F_{1}, F_{2}$ are the foci of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ , and $P$ is on the ellipse with $∠ F_{1} P F_{2} = 9 0^{\circ}$ . Find the area of $△ P F_{1} F_{2}$ .

Try this

Why this equation

From the definition to the standard equation

An ellipse is a squashed circle

If the string is too short

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Always putting the foci on the $x$-axis

Computing the ellipse's $c^2$ as $a^2+b^2$