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Question #40807

Difficulty: 8/10 | 🟢 Free

The Question

For a family of ellipses $E_e$ with eccentricity $e \in (0,1)$ , the perpendicular distance from the center to the tangent with slope $e$ is given by $p(e) = \sqrt{\frac{1-e}{2+e^2-e^4}}$ . Let $R(e)$ be the radius of the director circle of $E_e$ . If $f(e) = R(e^2)$ , evaluate $\lim_{e \to 1^-} \sqrt{1-e} f'(e)$

Options

A.
$-\frac{1}{\sqrt{2}}$
B.
$-\frac{1}{2\sqrt{2}}$
C.
$-\sqrt{2}$
D.
$0$

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

Director Circle Locus|Ellipse Auxiliary Circle|Chain Rule Inner Function Differentiability

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to link the parametric tangency condition of an ellipse to its director circle, while navigating a subtle Chain Rule trap.

Students must first correctly identify the distance formula to a tangent and use the eccentricity relation $b^2 = a^2(1-e^2)$ to find the semi-major axis $a$ .

The core trap lies in differentiating the composite function $f(e) = R(e^2)$ .

A common mistake is ignoring the inner function's derivative (the $2e$ factor from $e^2$ ), leading to the incorrect derivative $R'(e^2)$ and the trap answer $-\frac{1}{2\sqrt{2}}$ .

Furthermore, the limit evaluates the behavior as the ellipse approaches a parabolic transition ( $e \to 1$ ), where the director circle radius vanishes, testing their rigorous application of limits on indeterminate forms.

Quick Facts About This Question

Question Information

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