Let the maximum area of the triangle that can be inscribed in the ellipse x²/a² + y²/4 = 1, a 2, h

Q: Wait, why do we use parametric coordinates instead of just $(x, y)$?

Using $(x, y)$ forces you to deal with the constraint $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$ as a square root function, which makes differentiation messy. Parametric coordinates $(a\cos\theta, b\sin\theta)$ automatically satisfy the ellipse equation, reducing the problem to a single variable $\theta$.

Q: Why do we reject the solution $\cos\theta = 1$?

If $\cos\theta = 1$, then $x = a$. The vertices $B$ and $C$ would both be at $(a, 0)$, which is the same as vertex $A$. This collapses the triangle into a single point, giving an area of zero, which is the minimum, not the maximum.

Q: Is there a way to solve this without calculus?

Yes, you could use the AM-GM inequality if you manipulate the area expression into a product of terms whose sum is constant. However, calculus is a more universal tool that works for any function, making it a safer bet for JEE Advanced.

Question

Let the maximum area of the triangle that can be inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{4} = 1, a > 2$, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be $6\sqrt{3}$. Then the eccentricity of the ellipse is :

Options:

(A) $\frac{\sqrt{3}}{2}$

(B) $\frac{1}{2}$

(C) $\frac{1}{\sqrt{2}}$

(D) $\frac{\sqrt{3}}{4}$

Sigma Percentile · Accepted Answer

### The Geometry of the Ellipse: A Playground for Optimization

Imagine you are standing on a coordinate plane, staring at an ellipse defined by the equation:

$$\frac{x^2}{a^2} + \frac{y^2}{4} = 1$$

It is a beautiful, symmetric shape, and we are tasked with inscribing a triangle inside it. One vertex is fixed at the end of the major axis, $(a, 0)$, and the opposite side is a vertical line.

This setup is a classic challenge that tests your ability to bridge the gap between geometry and calculus.

### The Power of Parametric Coordinates

When we deal with conics, the standard Cartesian coordinates $(x, y)$ can sometimes feel like a straightjacket. There is a secret weapon in our toolkit: parametric coordinates.

By defining any point on the ellipse as $(a\cos	heta, b\sin	heta)$, we automatically satisfy the ellipse equation. Here, $b = \sqrt{4} = 2$.

So, our vertices $B$ and $C$ become $(a\cos	heta, 2\sin	heta)$ and $(a\cos	heta, -2\sin	heta)$. The side $BC$ is parallel to the $y$-axis because both points share the same $x$-coordinate, $a\cos	heta$.

### Building the Area Function

The base $BC$ is the vertical distance between $2\sin	heta$ and $-2\sin	heta$, which is $4\sin	heta$. The height is the horizontal distance from the vertex $A(a, 0)$ to the line $x = a\cos	heta$.

This distance is simply $a - a\cos	heta = a(1 - \cos	heta)$. With the base and height in hand, the area $A$ of the triangle is given by the formula $A = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.

Substituting our expressions, we get:

$$A(	heta) = \frac{1}{2} \cdot (4\sin	heta) \cdot a(1 - \cos	heta) = 2a(1 - \cos	heta)\sin	heta$$

This is our area function, a function of a single variable $	heta$.

### The Calculus of Optimization

To find the maximum area, we need to find where the derivative $\frac{dA}{d	heta}$ is zero. Using the product rule on $A(	heta) = 2a(1 - \cos	heta)\sin	heta$:

$$\frac{dA}{d	heta} = 2a [\sin^2	heta + (1 - \cos	heta)\cos	heta]$$

This simplifies to $2a [\sin^2	heta + \cos	heta - \cos^2	heta]$. Using the identity $\sin^2	heta = 1 - \cos^2	heta$, we obtain:

$$2a [1 - \cos^2	heta + \cos	heta - \cos^2	heta] = 2a [1 + \cos	heta - 2\cos^2	heta]$$

Setting this to zero gives us the quadratic $2\cos^2	heta - \cos	heta - 1 = 0$. Factoring this, we find $(2\cos	heta + 1)(\cos	heta - 1) = 0$.

We reject $\cos	heta = 1$ because it leads to a degenerate triangle. Thus, $\cos	heta = -\frac{1}{2}$, which corresponds to $	heta = \frac{2\pi}{3}$.

### The Final Reveal

With $\cos	heta = -\frac{1}{2}$, we find $\sin	heta = \frac{\sqrt{3}}{2}$. Substituting these into our area function:

$$A_{max} = 2a\left(1 - \left(-\frac{1}{2}ight)ight)\left(\frac{\sqrt{3}}{2}ight) = 2a\left(\frac{3}{2}ight)\left(\frac{\sqrt{3}}{2}ight) = \frac{3\sqrt{3}}{2}a$$

We are given that the maximum area is $6\sqrt{3}$. Equating $\frac{3\sqrt{3}}{2}a = 6\sqrt{3}$, we find **$a = 4$**.

Finally, the eccentricity $e$ is given by:

$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{16}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$

We have successfully navigated the problem, proving that with the right tools, even the most complex geometry becomes a clear, logical path.

Select Answer:

Visualized Solution

Visualizing the Ellipse

Defining Triangle Vertices

Calculating Base and Height

Formulating the Area Function

Differentiating for Maxima

Simplifying the Derivative

Solving the Quadratic Equation

Finding Maximum Area in terms of 'a'

Solving for 'a'

Calculating Eccentricity

Summary and Takeaway

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Geometry of the Ellipse

The Power of Parametric Coordinates

Building the Area Function

The Calculus of Optimization

The Final Reveal

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