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Question #40782

Difficulty: 8/10 | 🟢 Free

The Question

Consider the line $L$ given by the equation $(a^2 - 1)x - 2ay + (a^2 + 1)d = 0$ , where $a > 1$ and $d > 0$ . If the equation of $L$ is expressed in the normal form $x \cos \alpha + y \sin \alpha = p$ , where $p > 0$ and $\alpha \in [0, 2\pi)$ , then the ordered pair $(p, \alpha)$ is:

Options

A.
$(d, 2\tan^{-1}a)$
B.
$(d, \pi + 2\tan^{-1}a)$
C.
$(d, 2\pi - 2\tan^{-1}a)$
D.
$(d, \pi - 2\tan^{-1}a)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Normal Form Parameters|The Quadrant Adjustment Logic|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the normal form of a straight line, specifically the non-negotiable constraint that the perpendicular distance $p$ must be strictly positive.

Students often blindly divide by the magnitude of the normal vector without checking the sign of the constant term.

If they fail to multiply the equation by $-1$ to make the right-hand side positive, they will derive incorrect signs for $\cos \alpha$ and $\sin \alpha$ , effectively rotating the normal vector by $\pi$ radians (leading to the trap option $\pi + 2\tan^{-1}a$ ).

Furthermore, the problem requires mapping the algebraic coefficients to inverse trigonometric functions, forcing students to verify that the resulting angle $2\tan^{-1}a$ naturally falls into the correct quadrant (the second quadrant, since $a > 1$ ) without requiring additional phase shifts to satisfy $\alpha \in [0, 2\pi)$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Normal Form Parameters
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free