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Question #40818

Difficulty: 8/10 | 🟢 Free

The Question

Let $H$ be the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 1$ . A point $P(t)$ is defined as $P(t) = \left( \frac{3(2t + \cos t)}{e^{2t} - 1}, \frac{4(2t + \cos t)}{e^{2t} - 1} \right)$ for $t > 0$ . Let $L_t$ be the polar of $P(t)$ with respect to $H$ . As $t \to 0^+$ , the line $L_t$ approaches a limiting line $L$ . Find the perpendicular distance from the point $A(3, -1)$ to the line $L$ .

Options

A.
3
B.
$\frac{3}{5}$
C.
0
D.
$\frac{12}{5}$

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

Polar and Pole Duality|Asymptote Geometry|The L'Hôpital's Rule Misuse

Why This Question Is Tricky

Common Mistakes

This question tests the duality of poles and polars as the pole approaches infinity along an asymptote, combined with a lethal calculus trap.

Students must find the limit of the polar line equation.

The trap lies in evaluating $\lim_{t \to 0} \frac{e^{2t} - 1}{2t + \cos t}$ .

Because it visually resembles standard indeterminate forms, students mechanically apply L'Hôpital's rule without verifying the $0/0$ constraint, yielding a false limit of $1$ and an incorrect line $4x - 3y = 12$ .

Recognizing that the limit is simply $0/1 = 0$ reveals the geometric truth: the polar of a point approaching infinity along an asymptote is the asymptote itself.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Conic Sections - Class 11
Concept	Polar and Pole Duality
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free