The centre of an ellipse is at (0,0), major axis is on the y-axis. If the ellipse passes through (3,2) and (1,6), then what is its eccentricity?

Explanation

Let the equation of the ellipse be \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 (major axis on y-axis, so a \gt b). Passing through (3,2) gives \frac{9}{b^2} + \frac{4}{a^2} = 1. Passing through (1,6) gives \frac{1}{b^2} + \frac{36}{a^2} = 1. Multiplying the second by 9 and subtracting the first yields \frac{320}{a^2} = 8 \implies a^2 = 40. Substituting back gives \frac{1}{b^2} + \frac{36}{40} = 1 \implies \frac{1}{b^2} = \frac{1}{10} \implies b^2 = 10. The eccentricity is e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{10}{40}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}.

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