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Question #40802

Difficulty: 8/10 | 🟢 Free

The Question

Let $E$ be an ellipse whose auxiliary circle has an area of $16\pi$ and the distance between its directrices is $16$ . Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^2 - \frac{a}{2}x + a^2 e^2 = 0$ , where $a$ is the semi-major axis and $e$ is the eccentricity of $E$ . A sequence is defined as $T_k = \alpha^k + \beta^k$ for integers $k \ge 0$ . Find the value of $\frac{1}{2^{11}} \sum_{k=0}^{10} \binom{10}{k} T_k T_{10-k}$

Options

A.
-511
B.
-512
C.
512
D.
513

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

Ellipse Auxiliary Circle|Quadratic Roots in Sequences|Binomial Coefficient Summation Properties

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of coordinate geometry, complex numbers, and binomial summation identities.

The primary trap lies in the expansion of the product $T_k T_{n-k}$ .

Students often erroneously assume the summation distributes linearly without cross-terms, calculating only $\sum \binom{n}{k}(\alpha^n + \beta^n) = 2^n T_n$ , which leads directly to the lethal distractor -512.

Furthermore, evaluating $T_{10}$ requires transitioning the roots of the quadratic into Euler's polar form; attempting to expand $(1 \pm i\sqrt{3})^{10}$ via the standard binomial theorem is computationally heavy and prone to sign errors (leading to the distractors 512 and 513), testing the student's algebraic maturity and precision.

Quick Facts About This Question

Question Information

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