Let $E_1=\frac{x^2}{9}+\frac{y^2}{4}=1$ and $E_2=\frac{x^2}{d^2}+\frac{y^2}{b^2}=1$ be two ellipses and…
Let $E_1=\frac{x^2}{9}+\frac{y^2}{4}=1$ and $E_2=\frac{x^2}{d^2}+\frac{y^2}{b^2}=1$ be two ellipses and $\mathrm{R}$ be a rectangle with sides parallel to the coordinate axes. Let $E_1$ be inscribed ellipse in $\mathrm{R}$ and $E_2$ be circumscribed ellipse on R. If $E_2$ passes through $(0,4)$ then
$a=4, b=2 \sqrt{3}$
$a=12, b=16$
$a=16, b=16$
$a=2 \sqrt{3}, b=4$
Solution
Acc. to question.
$\begin{aligned} & E_1=\frac{x^2}{g}+\frac{y^2}{4}=1 \Rightarrow a=3, b=2 \\ & E_2=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\end{aligned}$
$\because \mathrm{E}_2$, passes through $(0,4)$
$\Rightarrow b= \pm 4$ for $E_2$
from figure $C=(3,2)$
$\mathrm{a}^2=12$ or $\mathrm{a}=2 \sqrt{3}$
