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Question #41170

Difficulty: 8/10 | 🟢 Free

The Question

Let $m > 0$ be the real root of the equation $\sin^{-1}\left(\frac{2m}{1+m^2}\right) - \cos^{-1}\left(\frac{1-m^2}{1+m^2}\right) + \tan^{-1}\left(\frac{2m}{1-m^2}\right) = -\frac{2\pi}{3}$ Let $L_1$ be the tangent to the ellipse $\frac{x^2}{4} + \frac{y^2}{13} = 1$ having slope $m$ and a positive $y$ -intercept. Let $L_2$ be the tangent to the circle $x^2 + y^2 - 2\sqrt{3}x - 4y - 9 = 0$ that is parallel to $L_1$ and furthest from it. The distance between $L_1$ and $L_2$ is:

Options

A.
4
B.
5
C.
6
D.
7

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

Variable Transformation Subsitution|Condition of Tangency|Distance Between Parallel Lines

Why This Question Is Tricky

Common Mistakes

The question tests the domain-restricted properties of Inverse Trigonometric Functions (ITFs) integrated with coordinate geometry.

A common trap is blindly applying the identity $\sin^{-1}(\frac{2x}{1+x^2}) = 2\tan^{-1}x$ , which is only valid for $|x| \le 1$ .

Doing so yields an extraneous negative slope $m = -\sqrt{3}$ , which coincidentally makes the line $L_1$ pass exactly through the circle's center, leading to an incorrect maximum distance of 4.

By enforcing $m > 0$ and requiring the correct piecewise ITF definitions for $m > 1$ , the student finds the true slope $m = \sqrt{3}$ .

The secondary trap involves correctly applying the distance formula between parallel lines while managing the signs of the constants.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Variable Transformation Subsitution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free