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Question #40812

Difficulty: 8/10 | 🟢 Free

The Question

Let C1C_1 be the conic given by 5x2+6xy+5y2=205x^2 + 6xy + 5y^2 = 20. Let P(h,k)P(h,k) be a point on the conic C2:5x2+6xy+5y2=λC_2: 5x^2 + 6xy + 5y^2 = \lambda, where λ>20\lambda > 20. The tangents drawn from PP to C1C_1 touch C1C_1 at points A(x1,y1)A(x_1, y_1) and B(x2,y2)B(x_2, y_2). If the absolute value of the determinant Δ=∣hk1x1y11x2y21∣\Delta = \begin{vmatrix} h & k & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} is equal to 320320, find the value of λ\lambda.

Options

  • A.

    40

  • B.

    80

  • C.

    100

  • D.

    160

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

Chord of Contact Logic|Determinant as Geometric Area/Volume|Singular vs. Non-Singular Threshold

Why This Question Is Tricky

Common Mistakes

This question tests the geometric interpretation of a determinant as twice the area of a triangle, combined with the properties of the chord of contact.

The core trap lies in the coordinate transformation required to simplify the area calculation.

Students often mistakenly use the determinant of the quadratic form matrix (det⁡(Q)=16\det(Q) = 16) as the area scaling factor, leading directly to the trap answer λ=40\lambda = 40.

The correct scaling factor requires normalizing the conic equation to the identity matrix, yielding det⁡(M)=5\det(M) = 5.

Furthermore, the condition λ>20\lambda > 20 ensures the non-singular case, as λ=20\lambda = 20 would make PP lie on C1C_1, degenerating the triangle to a line (Area = 0) and making the determinant singular.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterConic Sections - Class 11
ConceptChord of Contact Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree