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Question #41534

Difficulty: 8/10 | 🟢 Free

The Question

Let $E$ be the ellipse $x^2 + 3y^2 = 12$ . The polar of the point $P(6, 2)$ with respect to $E$ is the line $L$ . Let $L$ be represented by the equation $f(x,y) = 0$ , where $f(x,y)$ is a linear polynomial in $x$ and $y$ with coprime integer coefficients such that $f(0,0) > 0$ .

Consider the region $\mathcal{R} = \{ (x,y) \in \mathbb{R}^2 : x^2 + 3y^2 \le 12 \text{ and } f(x,y) \ge 0 \}$ .

The area of the region $\mathcal{R}$ is:

Options

A.
$\frac{8\pi\sqrt{3}}{3} + 3$
B.
$\frac{4\pi\sqrt{3}}{3} - 3$
C.
$\frac{8\pi\sqrt{3}}{3} - 3$
D.
$\frac{4\pi\sqrt{3}}{3} + 3$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

The Sign-Consistency Principle|Ellipse Auxiliary Circle|Polar and Pole Duality|Symmetric vs Non-symmetric form

Why This Question Is Tricky

Common Mistakes

This question tests the Sign-Consistency Principle and the Symmetric vs Non-symmetric form trap in coordinate geometry.

Students often blindly apply the $T=0$ formula to find the polar and immediately use it as the boundary function $f(x,y)$ without normalizing it to satisfy the given condition $f(0,0) > 0$ .

If they use $f(x,y) = x+y-2$ , the inequality $f(x,y) \ge 0$ incorrectly directs them to the smaller segment of the ellipse.

By forcing the non-symmetric form $2-x-y$ , the inequality flips, requiring the area of the larger segment.

Furthermore, it demands the use of the auxiliary circle transformation to efficiently calculate the area of the elliptical segment, a hallmark of JEE Advanced problems.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	The Sign-Consistency Principle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free