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Question #73877

Difficulty: 8/10 | 🟢 Free

The Question

Consider the family of conics $C_\lambda : \frac{x^2}{\lambda} + \frac{y^2}{1-\lambda} = 1$ , where $\lambda \in \mathbb{R} \setminus \{0, 1\}$ . Let $D_\lambda$ be the director circle of $C_\lambda$ . Let $\Gamma_\lambda$ be the locus of points $P$ such that the chord of contact of tangents from $P$ to $C_\lambda$ subtends a right angle at the origin. Let $S$ be the set of all $\lambda$ for which both $D_\lambda$ and $\Gamma_\lambda$ are real curves containing at least one point. For $\lambda \in S$ , let $e_1$ and $e_2$ be the eccentricities of $C_\lambda$ and $\Gamma_\lambda$ respectively. Which of the following is true?

Options

A.
$S = (0, 1)$ and $e_2^2 = e_1^2(2 - e_1^2)$
B.
$S = (0, \infty) \setminus \{1\}$ and $e_2^2 = e_1^2(2 - e_1^2)$
C.
$S = (0, 1)$ and $e_2^2 = \frac{2e_1^2}{1 + e_1^2}$
D.
$S = \mathbb{R} \setminus \{0, 1\}$ and $e_2^2 = \frac{e_1^2}{2 - e_1^2}$

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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

Position Vector Origin Dependence|Chord of Contact Logic|Director Circle Locus|Compound Function Boundaries

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the 'Compound Function Boundary' trap coupled with 'Chord of Contact Logic'.

Students often mistakenly assume the director circle of a hyperbola might not exist, but here $D_\lambda$ is always $x^2+y^2=1$ .

The true domain restriction comes from the locus $\Gamma_\lambda$ , where the homogenization process yields a RHS of $\frac{1}{\lambda(1-\lambda)}$ .

For the locus to be real, this must be positive, restricting $\lambda \in (0,1)$ .

Furthermore, calculating the eccentricity requires careful tracking of major/minor axes depending on $\lambda$ , but the algebraic relation $e_2^2 = e_1^2(2-e_1^2)$ elegantly unifies both cases, rewarding pure symbolic manipulation.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Position Vector Origin Dependence
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free