Math · Precalculus · Worked example

Find the vertices and foci of an ellipse

Find the vertices, co-vertices and foci of the ellipse 9x² + 4y² = 36.

9 x^{2} + 4 y^{2} = 36

Divide to get 1 on the right

Divide every term by 36.

\begin{array}{l} 9 x^{2} + 4 y^{2} = 36 \\ \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1 \end{array}

Find a and b

The larger denominator, 9, is under y², so the major axis is vertical: a² = 9 and b² = 4, so a = 3 and b = 2. The vertices are (0, ±3) and the co-vertices (±2, 0).

Find the foci

For an ellipse c² = a² − b² = 9 − 4 = 5, so c = √5 ≈ 2.24. The foci lie on the major axis at (0, ±√5).

c = \sqrt{9 - 4} = \sqrt{5} \approx 2.24

Check a vertex and the focal sum

The vertex (0, 3) satisfies the equation. The co-vertex (2, 0) is √(4 + 5) = 3 from each focus, and the two distances add to 2a = 6.

\begin{array}{l} \frac{0^{2}}{4} + \frac{3^{2}}{9} = 1 \\ 2 \sqrt{4 + 5} = 6 \end{array}

Result

Vertices (0, ±3), co-vertices (±2, 0), foci (0, ±√5) ≈ (0, ±2.24).

Your turn

Find the vertices and foci of x²/25 + y²/16 = 1.

Show the answer and explanation

Vertices (±5, 0); foci (±3, 0).

The larger denominator, 25, is under x², so the major axis is horizontal with a = 5 and b = 4. Then c² = 25 − 16 = 9, so c = 3.

\sqrt{25 - 16} = 3

Keep exploring

In Graph, the ellipse is drawn as two halves, y = ±(3/2)√(4 − x²). Change the 3/2 to 1 in both and the ellipse becomes the circle x² + y² = 4.

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