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ઉપવલય ${E_1}\,\,:\,\,\frac{{{x^2}}}{9}\,\, + \;\,\frac{{{x^2}}}{4}\, = \,\,1$એ લંબચોરસ $R$ કે જેની બાજુઓ યામાક્ષોને સમાંતર હોય તેની અંદર આવેલ છે બીજુ ઉપવલય $E_2\ (0, 4)$ તો ઉપવલય $E_2$ ની ઉત્કેન્દ્રતા :
$\frac{{\sqrt 2 }}{2}$
$\frac{{\sqrt 3 }}{2}$
$1/2$
$3/4$
Solution
The ellipse $E_1$ touches $x$ axis at $(\pm 3,0)$ and $y$ axis at $(0, \pm 2)$
Since the ellipse is inscribed in rectangle $R$ whose sides are parallel to the coordinate axis, the vertices of rectangle are $(\pm 3, \pm 2)$
Let the equation of ellipse $E_2$ be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
The ellipse circumscribes the rectangle $R$, so the vertices of rectangle lies on ellipse $E _2$
Therefore we get $\frac{9}{a^2}+\frac{4}{b^2}=1$
Given that ellipse $E_2$ passes through $(0,4)$
So we get $b =4$ and $a ^2=12$
We know that $a^2=b^2\left(1-e^2\right)$
$\Rightarrow e ^2=\frac{1}{4}$
$\Rightarrow e =\frac{1}{2}$