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?

We're lucky that the ellipse is centered at the origin. :)
In this case, the inscribed rectangle is also centered at the origin. If $P = (x, y)$ is the vertex of the inscribed rectangle at the first quadrant, then the smaller rectangle spanned by the origin and point $(x, y)$ is similar to the inscribed rectangle.
Each side of the inscribed rectangle is $2$ times that of the smaller rectangle. Hence, the area of the inscribed rectangle is $2^2 = 4$ times that of the smaller rectangle. The sides of the smaller rectangle are $x$ and $y$ respectively so we have $x \colon y = 3 \colon 2$. This means we have $2x = 3y$.
Plugging it into the equation of the ellipse, we have
\begin{align}
4x^2 + 9y^2 &= 144 \\
(2x)^2 + (3y)^2 &= 144 \\
(2x)^2 + (2x)^2 &= 144 \\
8x^2 &= 144 \\
x^2 &= 18 \Rightarrow x = \pm3\sqrt{2}
\end{align}
Since $(x, y)$ is in the first quadrant, we have $x = 3\sqrt{2}$ and hence $y = 2\sqrt{2}$. The area of the smaller rectangle is $xy = 12$, so the area of the inscribed rectangle is $4xy = 4 \cdot 12 = 48$.
Update
I have incorporated @zwim's comment into the answer. Yeah, it's more natural this way, thanks @zwim! :) Also I have corrected an error in the computation of $x$ and $y$ in my previous answer.
