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  • The ellipse <img alt="\mathrm{E_1: \frac{x^2}{9}+\frac{y^2}{4}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BE_1%3A%20%5Cfrac%7Bx%5E2%7D%7B9%7D&plus;%5Cfrac%7By%5E2%7D%7B

The ellipse \mathrm{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 \mathrm{E_2} passing through the point \mathrm{(0,4)} circumscribes the rectangle R. The eccentricity of the ellipse \mathrm{E_2} is

Option: 1

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


Option: 2

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


Option: 3

\frac{1}{2}


Option: 4

\frac{3}{4}


Answers (1)

best_answer

Let the required ellipse be \mathrm{\frac{x^2}{\alpha}+\frac{y^2}{\beta}=1}

It passes through \mathrm{(0,4) \Rightarrow \frac{16}{\beta}=1 \quad \therefore \beta=16}

It passes through \mathrm{(3,2) \Rightarrow \frac{9}{\alpha}+\frac{4}{\beta}=1 \Rightarrow \alpha=12}

Also \mathrm{\alpha=\beta\left(1-e^2\right) \Rightarrow 12=16\left(1-e^2\right)}

\mathrm{ \Rightarrow \frac{3}{4}=1-e^2 \Rightarrow e^2=\frac{1}{4} \quad \therefore \quad e=\frac{1}{2} }

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Riya

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