How to find area of rectangle inscribed in ellipse.

Question

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?

Answer 1

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.

Answer 2

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 $$\text{Length}\times \text{Width}=12\cos\theta\cdot 8\sin\theta$$ $$=48\sin2\theta$$ $$=48\sin\frac{\pi}{2}=\color{blue}{48\ \text{unit}^2}$$

Answer 3

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.

Answer 4

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 $$4\sqrt{18\cdot8}=48.$$

Answer 5

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.