Mathematics - Ellipse Question with Solution | TestHub

As rectangle $ABCD$ circumscribed the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1,\therefore A=(3,2)$
Let the ellipse circumscribing the rectangle $ABCD$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\mathrm{1...}(i)$ Given that ellipse (i) passes through $(a,4)$ $\therefore b^2=16$ Also ellipse (i) passes through $A(3,2)$ $\therefore \frac{9}{a^2}+\frac{4}{16}=1\Rightarrow a^2=12$ $\therefore e=\sqrt{1-\frac{12}{16}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$