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?

Consider the four corner/vertex points $(\pm 6\cos\theta, \pm 4\sin\theta)$ of rectangle lying on given ellipse: $4x^2+9y^2=144$
Now, the sides of the rectangle are length: $(2\cdot 6\cos\theta)$ & width $(2\cdot 4\sin\theta)$ which are in ratio $3:2$ as given in question therefore we have $$\frac{12\cos\theta}{8\sin\theta}=\frac32\implies \tan\theta=1\iff \theta=\frac{\pi}{4}$$
Now, the area of rectangle inscribed in given ellipse $$\text{Length}\times \text{Width}=12\cos\theta\cdot 8\sin\theta$$
$$=48\sin2\theta$$
$$=48\sin\frac{\pi}{2}=\color{blue}{48\ \text{unit}^2}$$
