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Question #41644

Difficulty: 8/10 | 🟢 Free

The Question

Let $E$ be the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$ in the $xy$ -plane ( $z=0$ ). A point $P$ moves along the portion of $E$ that lies within the region defined by $y^2 \le \frac{9}{4}x$ . Find the maximum distance from the point $P$ to the plane $\Pi: x - 2\sqrt{3}y + 3z = 5$ .

Options

A.
$\frac{4 + 3\sqrt{3}}{\sqrt{22}}$
B.
$\frac{5 + 2\sqrt{10}}{\sqrt{22}}$
C.
$\frac{3\sqrt{3} - 4}{\sqrt{22}}$
D.
$\frac{2\sqrt{10} - 5}{\sqrt{22}}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Distance from point to plane|Ellipse Auxiliary Circle|Extraneous Solution Rejection

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate 3D geometry with constrained trigonometric optimization, specifically targeting the 'Extraneous Solution Rejection' trap.

The 'Lethal Collision' occurs when students set the derivative of the distance function to zero and blindly accept the critical point ( $\tan\theta = -3$ ) to find the maximum distance, yielding the incorrect unconstrained maximum $\frac{5 + 2\sqrt{10}}{\sqrt{22}}$ .

Rigorous analysis requires mapping the geometric boundary of the parabolic region to the ellipse's eccentric angle, revealing that the valid domain is restricted to $\theta \in [-\pi/3, \pi/3]$ .

Since the critical point lies outside this physical boundary, the true maximum occurs at the domain's edge, forcing students to reject the extraneous root and evaluate the boundary conditions.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Distance from point to plane
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free