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Question #40830

Difficulty: 8/10 | 🟢 Free

The Question

A ray of light originating from the origin $O(0,0)$ travels in the $xy$ -plane and strikes the outer surface of a circular mirror whose equation is $x^2 + (y-2)^2 = 1$ . After reflection, the ray travels parallel to the x-axis. If the point of incidence $P(x, y)$ lies in the first quadrant, what is the exact value of its y-coordinate?

Options

A.
$\frac{15 - \sqrt{33}}{8}$
B.
$\frac{15 + \sqrt{33}}{8}$
C.
$\frac{11 - \sqrt{21}}{6}$
D.
$\frac{13 - \sqrt{33}}{8}$

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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

Reflective Property|The Sum-Difference Formula Constraints|Extraneous Solution Rejection

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to reject extraneous algebraic solutions based on geometric constraints, a hallmark of JEE Advanced.

While the vector reflection formula (or trigonometric angle bisector logic) correctly yields a quadratic with two real roots, one root corresponds to a point on the 'hidden' upper half of the circular mirror.

A ray from the origin would have to pass through the solid mirror to reach it.

Students who blindly accept both roots or guess the positive-looking one fall into the trap; rigorous domain restriction via tangent visibility ( $y \le 1.5$ ) is required to isolate the true physical point of incidence.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Conic Sections - Class 11
Concept	Reflective Property
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free