PrecalculusConic SectionsG-GPE.A.3

Ellipses

The sum of distances to two foci is constant; c² = a² − b².

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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.

Key idea

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 $y - k$.

Why it works

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 $a$.
Drag $P$ around the ellipse. The two distances change, but their sum stays $2a$. Change $a$ and $b$ to see the foci move.

Worked examples

Worked example: Reading an equation

Find the vertices and foci of $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$.
  1. The larger denominator, $25$, is under $x^2$: horizontal major axis, $a = 5$, $b = 3$.
  2. $c^2 = 25 - 9 = 16$, so $c = 4$.

Answer: Vertices $(\pm 5, 0)$; foci $(\pm 4, 0)$

Worked example: A tall ellipse

Find the foci of $\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1$.
  1. The larger denominator is under $y^2$: vertical major axis with $a = 5$, $b = 4$.
  2. $c^2 = 25 - 16 = 9$, $c = 3$, on the $y$-axis.

Answer: $(0, \pm 3)$

Worked example: Writing an equation

Write the ellipse centered at the origin with vertices $(\pm 13, 0)$ and foci $(\pm 12, 0)$.
  1. $a = 13$, $c = 12$, so $b^2 = a^2 - c^2 = 169 - 144 = 25$.
  2. The major axis is horizontal, so $a^2$ goes under $x^2$.

Answer: $\frac{x^2}{169} + \frac{y^2}{25} = 1$

Common mistake

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.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Also used in other ladders:

Sources and further reading

  • OpenStax Precalculus 2e, Section 10.1 The Ellipse
  • Common Core State Standards for Mathematics, G-GPE.A.3
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