Tie a loop of string around two pins and pull it tight with a pencil: as you move the pencil around, it draws an ellipse. The pins are the foci. Like circles, ellipses have tidy standard equations, and you can find the center by completing the square.
Idea clave
An ellipse is the set of points whose distances to two foci add to a constant, $2a$. Centered at the origin: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \ (a > b)$$ has a horizontal major axis, vertices $(\pm a, 0)$, co-vertices $(0, \pm b)$ and foci $(\pm c, 0)$, where $c^2 = a^2 - b^2$. If the larger denominator is under $y^2$, the major axis is vertical. With center $(h, k)$, replace $x$ by $x - h$ and $y$ by $y - k$.
Por qué funciona
At the vertex $(a, 0)$ the distances to the foci are $a - c$ and $a + c$, which add to $2a$, so the constant is $2a$. At the co-vertex $(0, b)$ the two distances are equal, so each is $a$; the right triangle with legs $b$ and $c$ and hypotenuse $a$ gives $b^2 + c^2 = a^2$. The foci sit inside the ellipse, so $c$ is smaller than $a$.
Worked examples
Ejemplo resuelto: Reading an equation
Find the vertices and foci of $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.
- The larger denominator, $25$, is under $x^2$: horizontal major axis, $a = 5$, $b = 3$.
- $c^2 = 25 - 9 = 16$, so $c = 4$.
Respuesta: Vertices $(\pm 5, 0)$; foci $(\pm 4, 0)$
Ejemplo resuelto: A tall ellipse
Find the foci of $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$.
- The larger denominator is under $y^2$: vertical major axis with $a = 5$, $b = 4$.
- $c^2 = 25 - 16 = 9$, $c = 3$, on the $y$-axis.
Respuesta: $(0, \pm 3)$
Ejemplo resuelto: Writing an equation
Write the ellipse centered at the origin with vertices $(\pm 13, 0)$ and foci $(\pm 12, 0)$.
- $a = 13$, $c = 12$, so $b^2 = a^2 - c^2 = 169 - 144 = 25$.
- The major axis is horizontal, so $a^2$ goes under $x^2$.
Respuesta: $\frac{x^2}{169} + \frac{y^2}{25} = 1$
Error común
Using $c^2 = a^2 + b^2$ (the hyperbola rule) for an ellipse, or reading the major axis from the order of the terms instead of the larger denominator. Checks: the foci of an ellipse are inside it, so $c < a$; and the major axis always goes with the bigger denominator.