In the Ellipse Shown Below: Read a, b, c, the Foci, and the Equation Off the Figure

The most famous ellipse behind every in the ellipse shown below question hangs in the sky, not in a textbook.

In the early 1600s, Kepler studied twenty years of Mars data and got a surprise: Mars does not circle the Sun. Its path is a squashed circle — an ellipse. A 1908 classic, A Scrap-book of Elementary Mathematics, records the law. Its one line: "Every planet moves in an elliptical orbit with the sun at one focus." That one orbit is the original in the ellipse shown below scene. Modern tests quote it directly: in the ellipse shown below, locate this, compute that.

An ellipse owns two special points, the foci. The Sun sits on one; the other stays empty. The marvel of the in the ellipse shown below orbit: wherever the planet goes, its two focal distances add to the same number. Every in the ellipse shown below skill starts at those two points, and the sum never budges.

That constant sum is the in the ellipse shown below definition. Call it $2a$. In the ellipse shown below, In our scale drawing, the orbit keeps the sum at 10, so $2a=10$ and $a=5$. That $a$ runs from center to the far end of the orbit: the semi-major axis.

Definition: an ellipse is the set of points whose distances to two fixed points (the foci) always add up to the constant $2a$.

One figure, every number visible — that is the engine of each in the ellipse shown below problem. The four numbers inside any in the ellipse shown below figure do all the work.

Coordinates turn the definition into the equation of an ellipse — the destination of every in the ellipse shown below solution. The in the ellipse shown below algebra places the foci at $F_1(-c,0)$ and $F_2(c,0)$. Write $PF_1+PF_2=2a$ with the distance formula, then isolate, square, and simplify. The in the ellipse shown below standard form appears, centered at the origin, foci on the x-axis:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad (a>b>0)$

Foci on the y-axis instead? In the ellipse shown below flip, the denominators trade places: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$. A 1917 Woods & Bailey text compresses the in the ellipse shown below reading rule: the larger denominator names the axis carrying the foci. That one line settles the first move of nearly every in the ellipse shown below problem.

The ellipse formula hides a right triangle

The three numbers inside the ellipse formula share one Pythagorean link — the backbone of every in the ellipse shown below computation:

$c^2=a^2-b^2$

A picture makes it click: the minor-axis endpoint $B$, the center $O$, and the focus $F$ form a right triangle. Its legs are $OB=b$ and $OF=c$; its hypotenuse is $BF=a$. The in the ellipse shown below triangle sits inside every drawing. Run the numbers on the in the ellipse shown below standard: $a=5$, $b=3$, so $c^2=25-9=16$ and $c=4$ — the foci land at $(\pm 4, 0)$.

A fourth number grades the flatness of any in the ellipse shown below drawing: the eccentricity.

Equations read one way; tests read the other. Below is what a typical in the ellipse shown below prompt looks like, and four steps take it apart. Learn how to graph an ellipse here, and every in the ellipse shown below figure turns transparent.

1. Spot the direction. The in the ellipse shown below figure stretches farther along one axis; the foci lie on that axis.
2. Read a and b. In the ellipse shown below figures, the far vertex gives $a$; the near vertex gives $b$. In equations, the larger denominator is $a^2$.
3. Compute c. The in the ellipse shown below numbers obey $c^2=a^2-b^2$; the foci sit on the major axis, a distance $c$ from the center.
4. Take the eccentricity. $e=c/a$; the in the ellipse shown below family keeps $e$ under 1, always.

The in the ellipse shown below figure hands you $a=5$ and $b=3$, so $c=\sqrt{25-9}=4$ and $e=4/5$. That is the entire in the ellipse shown below routine: direction, a, b, c. Run the four steps on every in the ellipse shown below figure you meet; the order never changes.

A compass finds the foci in seconds

Unsure where the foci go? A compass settles it in one arc. Pin the needle at the minor-axis endpoint $B$, open the radius to $a$, and swing. The two crossings on the major axis are the foci, because $OB=b$, $OF=c$, and the radius $BF=a$ is the hypotenuse. The dashed segment in the figure is that radius. No in the ellipse shown below drawing can hide its foci from the oldest trick in the toolkit for in the ellipse shown below problems.

Shape at a glance

A 1906 Wentworth algebra text put shape-reading into words. When $a$ and $b$ differ greatly, the ellipse stretches long and narrow. Nearly equal, the curve looks like a flattened circle. Exactly equal, it is a circle. Every in the ellipse shown below figure announces its shape the instant you compare a with b. That comparison is a one-second in the ellipse shown below check.

Problem: Put $4x^2+9y^2=36$ into standard form, then read off the major-axis direction, $a$, $b$, the foci, and the eccentricity. It is the in the ellipse shown below drill run in reverse: equation first, figure second.

Solution: Divide both sides by 36 — every in the ellipse shown below standardization starts there — to get $\frac{x^2}{9}+\frac{y^2}{4}=1$.

The larger denominator, 9, sits under $x^2$, so the in the ellipse shown below major axis runs along the x-axis. Read off $a=3$ and $b=2$.

Compute the foci: $c^2=a^2-b^2=9-4=5$, so $c=\sqrt{5}$ and the foci are $(\pm\sqrt{5}, 0)$.

Eccentricity: $e=\frac{\sqrt{5}}{3}\approx 0.745$.

Answer: major axis horizontal; $a=3$, $b=2$; foci $(\pm\sqrt{5}, 0)$; $e=\frac{\sqrt{5}}{3}$.

Run those numbers on a real in the ellipse shown below figure, and every label checks out.

Problem: An ellipse has vertices $(\pm 6, 0)$ and foci $(\pm 4, 0)$. Find its equation. This flips the usual in the ellipse shown below task: the graph arrives first, the equation second.

Solution: The in the ellipse shown below data hides in plain sight. The vertices hand you $a$: center to vertex is 6, so $a=6$. The foci hand you $c$: focus to center is 4, so $c=4$.

Use the Pythagorean link: $b^2=a^2-c^2=36-16=20$.

Substitute into the standard form: $\frac{x^2}{36}+\frac{y^2}{20}=1$.

Check: Substitute the vertex $(6,0)$: $\frac{36}{36}+0=1$ holds. Then $\sqrt{36-20}=4$, so the foci truly sit at $(\pm 4, 0)$. The in the ellipse shown below round-trip closes.

Answer: $\frac{x^2}{36}+\frac{y^2}{20}=1$.

Reverse reading carries half the in the ellipse shown below workload, so drill it in both directions.

Problem: An ellipse has eccentricity $e=\frac{1}{3}$; its foci are $(-1, 4)$ and $(7, 4)$. Find the equation. Off the origin, the in the ellipse shown below reading rules still hold.

Solution: The center is the midpoint of the foci: $\left(\frac{-1+7}{2}, \frac{4+4}{2}\right)=(3, 4)$.

Both foci sit at height 4, so the segment between them is horizontal. The in the ellipse shown below major axis runs parallel to the x-axis.

Each focus lies 4 units from the center, so $c=4$. From $e=\frac{c}{a}=\frac{1}{3}$, we get $a=12$.

Use the in the ellipse shown below Pythagorean link: $b^2=a^2-c^2=144-16=128$.

Substitute into the shifted standard form: $\frac{(x-3)^2}{144}+\frac{(y-4)^2}{128}=1$.

Check: Expand, and $8x^2+9y^2-48x-72y-936=0$ appears. Re-check the focus $(7,4)$: $\sqrt{(7-3)^2+0}=4=c$.

Answer: $\frac{(x-3)^2}{144}+\frac{(y-4)^2}{128}=1$, which expands to $8x^2+9y^2-48x-72y-936=0$.

The in the ellipse shown below habit survives every translation of the center.

(From Woods & Bailey, Analytic Geometry and Calculus, 1917, p. 77, Ex. 2 — numbers unchanged.)

1. Assuming the major-axis direction. You see $4x^2+2y^2=1$ and declare the foci sit on the x-axis. Rewrite it first, the in the ellipse shown below way: $\frac{x^2}{1/4}+\frac{y^2}{1/2}=1$. The $y^2$ denominator is larger, so the major axis is vertical and the foci sit on the y-axis. Direction is the first thing an in the ellipse shown below figure tests, and the verdict waits for standard form.

2. Writing the Pythagorean link as a sum. An ellipse obeys $c^2=a^2-b^2$. Writing $c^2=a^2+b^2$ borrows the hyperbola's rule — keep the two curves separate. The swap tops most in the ellipse shown below error lists.

3. Flipping the eccentricity fraction. The ratio is $e=\frac{c}{a}$, never $\frac{a}{c}$. An ellipse must keep $0