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Question #40829

Difficulty: 8/10 | 🟢 Free

The Question

Let $E$ be the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$ with foci $S_1$ and $S_2$ , where $S_1$ has a positive abscissa. A ray of light emanating from $S_1$ strikes the ellipse at a point $P$ and is reflected. The line containing the reflected ray intersects the line passing through $S_1$ and perpendicular to the tangent at $P$ at a point $Q$ . If the point $P$ is restricted to the region $x \ge 0$ and $y \ge 0$ , the locus of $Q$ is a curve of length $L$ . Find the value of $L$ .

Options

A.
$\frac{4\pi}{3}$
B.
$\frac{2\pi}{3}$
C.
$2\pi$
D.
$8\pi$

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

Reflective Property|Locus of a Moving Point|Domain Restrictions

Why This Question Is Tricky

Common Mistakes

This question tests the optical property of ellipses and the locus of the reflection of a focus across a tangent.

The core trap is the 'Domain Restriction'.

Students often correctly identify that the locus of $Q$ is the circle centered at the other focus with radius $2a$ and might hastily calculate the perimeter of the full circle ( $8\pi$ ) or assume a standard quadrant maps to a quarter circle ( $2\pi$ ).

However, restricting the point $P$ to the first quadrant non-linearly maps the angular domain of $Q$ .

The student must explicitly calculate the angular bounds of the vector $\vec{S_2 Q}$ to find that it subtends $\pi/3$ , not $\pi/2$ , making the locus a finite arc of length $4\pi/3$ .

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