How to find area of rectangle inscribed in ellipse. In an ellipse $4x^2+9y^2=144$ inscribed is a rectangle whose vertices lies on the ellipse and whose sides are parallel with the ellipse axis.
Longer side which is parallel to major axis, relates to the shorter sides as $3:2$. Find area of rectangle.
I can find the values of $a$ and $b$ as
$$\frac{4x^2}{144}+\frac{9y^2}{144}=1$$
$$\frac{x^2}{6^2}+\frac{y^2}{4^2}=1$$
Comparing with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, gives
$a=6$ & $b=4$. From here I have no idea how to solve further?

Let the top-right corner be at $(x,y)$. Squaring the aspect ratio, we have the system
$$\begin{cases}4x^2+9y^2=144,\\4x^2=9y^2\end{cases}$$
the solution of which is $x^2=18,y^2=8$.
Area $$4\sqrt{18\cdot8}=48.$$
