Ellipses and Hyperbolas from Their Foci

Two foci and a loop of string

Push two pins into a board, tie a string to both, and pull it tight with a pencil. Moving the pencil around keeps the string taut, so the two distances from the pencil to the pins always add to the string’s length. The curve it traces is an ellipse, and the pins are its foci.

That is the definition:

$$\text{an ellipse is every point whose distances to two foci add to } 2a$$

With foci at $(-3, 0)$ and $(3, 0)$ and a sum of $10$, the point $(3, 3.2)$ is $3.2$ from one focus and $6.8$ from the other, which add to $10$.

• the ellipse

Deriving the equation

Let $(x, y)$ be any point on this ellipse. The definition says

$$\sqrt{(x + 3)^2 + y^2} + \sqrt{(x - 3)^2 + y^2} = 10$$

Two square roots cannot be removed at once. Isolate one and square:

$$(x + 3)^2 + y^2 = 100 - 20\sqrt{(x - 3)^2 + y^2} + (x - 3)^2 + y^2$$

Expanding cancels the $x^2$, $y^2$ and $9$ on both sides:

$$12x - 100 = -20\sqrt{(x - 3)^2 + y^2} \quad\Rightarrow\quad 25 - 3x = 5\sqrt{(x - 3)^2 + y^2}$$

Square again:

$$625 - 150x + 9x^2 = 25x^2 - 150x + 225 + 25y^2$$ $$400 = 16x^2 + 25y^2 \quad\Rightarrow\quad \frac{x^2}{25} + \frac{y^2}{16} = 1$$

The general case runs the same way. With foci $(\pm c, 0)$ and sum $2a$:

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \qquad b^2 = a^2 - c^2$$

Here $a = 5$, $c = 3$ and $b = 4$: the ellipse reaches $5$ across and $4$ up.

Why squaring twice adds no false points

Squaring both sides can create extraneous solutions, so each squaring step here needs a reason to be safe.

The first squaring is safe because both sides were non-negative: one side is a distance, and the other is $10$ minus a distance that can never exceed $10$. The second is safe because $25 - 3x$ is positive for every $x$ between $-5$ and $5$, and the right side is a positive multiple of a distance.

Non-negative numbers with equal squares are equal, so each squared equation has exactly the same solutions as the one before it. The final equation describes the ellipse and nothing else.

The relation $b^2 = a^2 - c^2$ also has a picture. The top of the ellipse, $(0, b)$, is equally far from both foci, so each distance is $a$. That makes a right triangle with legs $c$ and $b$ and hypotenuse $a$.

The hyperbola

Change “add” to “differ” and the curve splits in two. A hyperbola is every point whose distances to the two foci differ by a constant $2a$.

With foci $(\pm 5, 0)$ and difference $6$, the same squaring twice gives

$$\frac{x^2}{9} - \frac{y^2}{16} = 1, \qquad b^2 = c^2 - a^2 = 25 - 9 = 16$$

• the hyperbola
• asymptotes

The point $(5, \tfrac{16}{3})$ is on the right branch: it is $\tfrac{16}{3}$ from the near focus and $\tfrac{34}{3}$ from the far one, and those differ by $6$.

Far from the center, the $1$ in the equation hardly matters, and the branches approach the lines $y = \pm\tfrac{b}{a}x = \pm\tfrac{4}{3}x$, the asymptotes.

	Ellipse	Hyperbola
definition	distances add to $2a$	distances differ by $2a$
equation	$\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1$	$\tfrac{x^2}{a^2} - \tfrac{y^2}{b^2} = 1$
foci	$(\pm c, 0)$, $c^2 = a^2 - b^2$	$(\pm c, 0)$, $c^2 = a^2 + b^2$
vertices	$(\pm a, 0)$	$(\pm a, 0)$

Worked examples

Common mistakes

Practice problems