Ellipse E₁: frac{x^2}{9}+frac{y^2}{4}=1 is inscribed in a rectangle R whose sides are parallel to

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10-2. Parabola, Ellipse, Hyperbola

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The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes.

Another ellipse $E _2$ passing through the point $(0,4)$ circumscribes the rectangle $R$.. The eccentricity of the ellipse $E _2$ is

A

$\frac{\sqrt{2}}{2}$

B

$\frac{\sqrt{3}}{2}$

C

$\frac{1}{2}$

D

$\frac{3}{4}$

(IIT-2012)

Solution

Let required ellipse is

$E_2: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

It passes thorugh $(0,4)$

$0+\frac{16}{b^2}=1 \quad \Rightarrow \quad b^2=16$

It also passes through $( \pm 3, \pm 2)$

$\frac{9}{a^2}+\frac{4}{b^2}=1 $

$\frac{9}{a^2}+\frac{1}{4}=1 $

$\frac{9}{a^2}=\frac{3}{4} \quad \Rightarrow \quad a^2=b^2\left(1-e^2\right) $

$\frac{12}{16}=1-e^2 $

$e^2=1-\frac{12}{16}=\frac{4}{16}=\frac{1}{4} $

$e=\frac{1}{2}$

Standard 11

Mathematics

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