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?
 A: 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.
A: 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 $$4\sqrt{18\cdot8}=48.$$
A: 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 $$\text{Length}\times \text{Width}=12\cos\theta\cdot 8\sin\theta$$
$$=48\sin2\theta$$
$$=48\sin\frac{\pi}{2}=\color{blue}{48\ \text{unit}^2}$$
A: Employing symbols for axes enables convenience with algebraic calculations. Given ellipse is in the form
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = k \tag1$$
with
$$ k=4, a=3,b=2 $$
The ellipse is intersected by a pair of straight lines
$$ \dfrac{y}{x}= \pm\dfrac{b}{a} \tag2$$
Solve 1) and 2)
$$ x_1= \dfrac{ka}{\sqrt2},\;y_1= \dfrac{kb}{\sqrt2}\;$$
Rectangle area is obtained by multiplying the above
$$   \dfrac{k^2ab}{2} \tag3 $$
calculates to 48 when given numerical values are plugged in.
A: Notice that everything is symmetric (specifically around both the x and y axes). So finding just one vertex is enough.
You know that the vertexes are on the ellipse, so their coordinates satisfy the defining equation of the ellipse. The 3:2 aspect ratio constraint also defines an equation the vertex must meet, so you have two independent equations of two unknowns. That should be enough to solve for the first vertex.
