Before You Begin

Question #40800

Difficulty: 8/10 | 🟢 Free

The Question

Let $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ( $a > b > 0$ ) be an ellipse with eccentricity $e$ . Let $P(ae, y_1)$ be a point on $E$ in the first quadrant. A matrix $M$ is defined as $M = \begin{pmatrix} a & 2y_1 \\ 2y_1 & a e^2 \end{pmatrix}$ Let $S$ be the set of all possible values of $e$ for which $M$ is a singular matrix. The sum of the elements of $S$ is:

Options

A.
$\frac{\sqrt{17}}{2}$
B.
$\frac{\sqrt{17}-1}{4}$
C.
$\frac{9}{4}$
D.
$\frac{\sqrt{17}+1}{4}$

Check Your Answer

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

Latus Rectum Geometry|Singular vs Non-Singular Matrices|Inverse Existence Criterion|Matrix Equality

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of matrix singularity and conic section constraints.

The core trap lies in solving the determinant equation $\det(M) = 0$ , which yields a degree-4 polynomial in $e$ .

Students will correctly find two positive roots for the eccentricity: $e = (\sqrt{17}-1)/4$ and $e = (\sqrt{17}+1)/4$ .

The lethal mistake is blindly summing these roots (yielding Option A) or using Vieta's formulas on $e^2$ (yielding Option C) without verifying the geometric constraint.

Because the curve is explicitly defined as an ellipse, the eccentricity must strictly satisfy $0 < e < 1$ .

The larger root corresponds to a hyperbola and must be rejected, leaving only a single valid element in the set.

Quick Facts About This Question

Question Information

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