Find the Area of the Largest Rectangle That Can Be Inscribed in the Ellipse x²/a² + y²/b² = 1 With Its Sides Parallel to the Coordinate Axes

A standard Class 12 maxima-and-minima problem, not tied to a specific exam paper.

Problem

Find the area of the largest rectangle that can be inscribed in the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, with its sides parallel to the coordinate axes.

Key insight. A rectangle inscribed symmetrically in the ellipse has one vertex at $(x,y)$ on the curve, with the rectangle’s area equal to $4xy$ by symmetry. Writing that vertex in the ellipse’s own parametric form $(a\cos\theta, b\sin\theta)$ turns the area into a single-variable trig expression that’s easy to maximise.

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Approach

Place one vertex of the rectangle at a general point on the ellipse using the parametric form $(a\cos\theta, b\sin\theta)$. By symmetry, the rectangle’s full width is $2a\cos\theta$ and height is $2b\sin\theta$, so its area is $4ab\sin\theta\cos\theta$. Rewrite this using the double-angle identity and maximise over $\theta$.

Solution

Step 1 — Parametrise a vertex of the rectangle on the ellipse

$$x = a\cos\theta, \qquad y = b\sin\theta$$

Step 2 — Express the rectangle’s area

By symmetry about both axes, the rectangle’s corners are at $(\pm x,\pm y)$, so its width is $2x$ and height is $2y$:

$$A = (2x)(2y) = 4xy = 4ab\sin\theta\cos\theta$$

Step 3 — Rewrite using the double-angle identity

$$A = 2ab\left(2\sin\theta\cos\theta\right) = 2ab\sin2\theta$$

Step 4 — Maximise over θ

Since $\sin2\theta$ has a maximum value of $1$ (attained at $2\theta=90°$, i.e. $\theta=45°$, a valid choice):

$$A_{\max} = 2ab\times1 = 2ab$$

Step 5 — Verify using calculus (optional check)

Alternatively, treating $A(x)=4x\cdot b\sqrt{1-x^2/a^2}$ as a function of $x$ and setting $\dfrac{dA}{dx}=0$ gives $x=\dfrac{a}{\sqrt2}$, and correspondingly $y=\dfrac{b}{\sqrt2}$:

$$A_{\max} = 4\times\frac{a}{\sqrt2}\times\frac{b}{\sqrt2} = 4\times\frac{ab}{2} = 2ab$$

Both methods agree.

Answer

Common mistakes

• Forgetting the factor of 4 from symmetry. The point $(x,y)$ on the ellipse gives only one corner of the rectangle — the full rectangle spans from $-x$ to $x$ and $-y$ to $y$, so its area is $4xy$, not just $xy$ or $2xy$.
• Maximising $xy$ directly using Lagrange multipliers or substitution without the parametric shortcut. This works too, but the parametric substitution turns a constrained two-variable optimisation into an unconstrained single-variable trig maximisation — a significant simplification worth recognising for any “extremum on an ellipse” problem.

Practise next

• Find the area of the largest rectangle that can be inscribed in the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$, using the same parametric approach, and confirm your numeric answer matches $2ab$ with $a=4,b=3$.