Area of the greatest rectangle that can be inscribed in the ellipse x²/a² + y²/b² = 1 is (A) 2ab (B)

Q: Wait, why do we use $2a \cos \theta$ and $2b \sin \theta$ for the dimensions? Shouldn't it just be $a \cos \theta$ and $b \sin \theta$?

That is a fantastic question! Remember, the point $(a \cos \theta, b \sin \theta)$ is just one vertex in the first quadrant. Because the rectangle is centered at the origin and symmetric, the rectangle extends from $-a \cos \theta$ to $+a \cos \theta$ along the x-axis, making the total length $2a \cos \theta$. Similarly, it extends from $-b \sin \theta$ to $+b \sin \theta$ along the y-axis, making the total height $2b \sin \theta$. Always account for the full span across the axes!

Q: Why can't we just use $x$ and $y$ and maximize $A = 4xy$ subject to the constraint $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$?

You absolutely can! That is a perfectly valid approach using Lagrange multipliers or substitution. However, using parametric coordinates $(a \cos \theta, b \sin \theta)$ is often much faster because it automatically satisfies the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ without needing to deal with square roots or complex derivatives. It turns a constrained optimization problem into a simple trigonometric maximization.

Q: Is there a physical intuition for why the maximum area occurs at $\theta = 45^\circ$?

Think about the balance between the length and the width. If you make the rectangle very thin and wide (near $\theta = 0$), the height is near zero. If you make it very tall and narrow (near $\theta = 90^\circ$), the width is near zero. The maximum area occurs when you find the perfect 'sweet spot' where the product of the dimensions is maximized, which happens when the contributions from the sine and cosine components are balanced, specifically at $\theta = 45^\circ$.

Question

Area of the greatest rectangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

Options:

(A) 2ab

(B) ab

(C) $\sqrt{ab}$

(D) a/b

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a perfectly drawn ellipse, defined by the elegant equation:

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

It is a shape of pure symmetry, a flattened circle that holds secrets of planetary orbits and architectural beauty. Our mission is to find the greatest rectangle that can be cradled within this curve, seeking the optimal balance between width and height.

### The Power of Symmetry

To begin, let us visualize the rectangle. Because the ellipse is symmetric about both the $x$-axis and the $y$-axis, the largest rectangle must also be centered at the origin.

If we place one vertex in the first quadrant at a point $P(x, y)$, the symmetry dictates that the other three vertices must be at $(-x, y)$, $(-x, -y)$, and $(x, -y)$. This means our rectangle has a total length of $2x$ and a total height of $2y$. Our goal is to maximize the area:

$$ A = (2x)(2y) = 4xy $$

### Embracing Parametric Elegance

While we could use Cartesian coordinates, the constraint $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ makes direct substitution messy. Instead, let us invite trigonometry to the party.

We can represent any point on the ellipse using the parametric coordinates $x = a \cos 	heta$ and $y = b \sin 	heta$, where $	heta$ is the eccentric angle. This substitution is a stroke of genius because it satisfies the ellipse equation automatically:

$$ \frac{(a \cos 	heta)^2}{a^2} + \frac{(b \sin 	heta)^2}{b^2} = \cos^2 	heta + \sin^2 	heta = 1 $$

Now, our rectangle's dimensions become $2a \cos 	heta$ and $2b \sin 	heta$.

### The Area Function

With these dimensions, the area $A$ becomes a function of the single variable $	heta$:

$$ A = (2a \cos 	heta) \cdot (2b \sin 	heta) $$

Simplifying this, we get $A = 4ab \sin 	heta \cos 	heta$. We recognize the trigonometric identity $2 \sin 	heta \cos 	heta = \sin 2	heta$. By rewriting our area formula, we obtain:

$$ A = 2ab \sin 2	heta $$

Look at the elegance of this result! The area is directly proportional to $\sin 2	heta$. Since $a$ and $b$ are fixed constants, the area is entirely controlled by the sine function.

### The Peak of the Curve

We know that the maximum value of the sine function is $1$. This occurs when the argument of the sine function is $\frac{\pi}{2}$.

Therefore, we set $2	heta = \frac{\pi}{2}$, which gives us $	heta = \frac{\pi}{4}$, or $45^\circ$. When $	heta = 45^\circ$, $\sin 2	heta = 1$, and the area reaches its absolute peak. Substituting this back into our equation, we find the maximum area:

$$ A_{max} = 2ab(1) = 2ab $$

### A Final Reflection

We started with a complex geometric constraint and, through the power of parametric coordinates and trigonometric identities, reduced it to a simple maximization of a sine wave. The result, **$2ab$**, is beautifully simple.

It tells us that for any ellipse, the greatest inscribed rectangle will always have an area exactly equal to twice the product of its semi-axes. Keep this result in your toolkit—it is a classic JEE Advanced gem that highlights how symmetry and substitution can turn a daunting problem into a moment of pure mathematical clarity.

Animated Solution for Mathematics - Differentiation: Area of the greatest rectangle that can be inscribed in the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is

Select Answer:

Visualized Solution

Equation of the Ellipse

Parametric Coordinates

Dimensions of the Rectangle

Setting up the Area Formula

Rearranging the Terms

Simplifying with Trigonometry

Maximizing the Area

Final Calculation

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Power of Symmetry

Embracing Parametric Elegance

The Area Function

The Peak of the Curve

A Final Reflection

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