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Question #40836

Difficulty: 8/10 | 🟢 Free

The Question

A circle C:x2+y2−10x−2y=0C: x^2 + y^2 - 10x - 2y = 0 intersects the parabola y2=4xy^2 = 4x at the origin OO and three other distinct points P,Q,RP, Q, R. Let t1,t2,t3t_1, t_2, t_3 be the parameters of P,Q,RP, Q, R respectively, where the coordinates of any point on the parabola are given by (t2,2t)(t^2, 2t). Let M=(t1t2t3t2t3t1t3t1t2)M = \begin{pmatrix} t_1 & t_2 & t_3 \\ t_2 & t_3 & t_1 \\ t_3 & t_1 & t_2 \end{pmatrix}. If B=adj(M−2I)B = \text{adj}(M - 2I), where II is the 3×33 \times 3 identity matrix, find the value of ∣B∣|B|.

Options

  • A.

    784

  • B.

    576

  • C.

    400

  • D.

    1024

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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

Intersection of Conics|Normal to Conic|Properties of Transpose Determinant|Adjoint-Determinant Relation

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of coordinate geometry (intersection of conics) with advanced matrix determinant properties.

The primary trap is the brute-force calculation: students might attempt to explicitly solve the cubic t3−6t−4=0t^3 - 6t - 4 = 0 (which has irrational roots) and substitute them into the 3×33 \times 3 matrix, leading to a massive time-sink and high probability of algebraic error.

The elegant, intended pathway requires recognizing that the matrix MM is a symmetric circulant-like matrix whose characteristic polynomial can be expressed entirely in terms of the elementary symmetric polynomials of its entries.

By leveraging Vieta's formulas and the adjoint-determinant relation ∣adj(A)∣=∣A∣n−1|adj(A)| = |A|^{n-1}, the problem bypasses raw coordinate computation entirely.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterConic Sections - Class 11
ConceptIntersection of Conics
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree