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Question #40841

Difficulty: 8/10 | 🟢 Free

The Question

Consider the system of linear equations in variables $p, q,$ and $r$ : $\lambda p + q + r = 1$ $p + \lambda q + r = \lambda$ $p + q + \lambda r = \lambda^2$ where $\lambda \in \mathbb{R}$ . Let $S$ be the set of all values of $\lambda$ for which the system has a unique solution $(p, q, r)$ such that the equation $\frac{x^2}{q} - \frac{y^2}{r} = 1$ represents a valid hyperbola. If for some $\lambda \in S$ , the area of the triangle formed by any tangent to this hyperbola and its asymptotes is $\frac{2}{3}$ sq. units, then the value of $5|\lambda|$ is:

Options

A.
7
B.
5
C.
14
D.
2

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Asymptote Geometry|Condition of Tangency|Cramer's Rule Constraints

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of Cramer's Rule with coordinate geometry.

The primary trap lies in the absolute value equation $|\lambda+1|/(\lambda+2) = 2/3$ , which yields two mathematical roots: $\lambda = 1$ and $\lambda = -7/5$ .

Students often blindly accept $\lambda = 1$ because it is a positive integer and seems 'cleaner'.

However, substituting $\lambda = 1$ back into the system's determinant yields $D = 0$ , violating the 'unique solution' constraint established by Cramer's Rule.

Additionally, the problem requires the knowledge that the area of the triangle formed by a hyperbola's tangent and its asymptotes is invariant and equals $ab$ .

Quick Facts About This Question

Question Information

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