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Question #41575

Difficulty: 8/10 | 🟢 Free

The Question

Let $A(u)$ be the area of the smaller region bounded by the ellipse $x^2 + 4y^2 = 4$ and the line $x+y=u$ , where $0 \leq u \leq \sqrt{5}$ . It is given that the rate of change of this area with respect to $u$ is $\frac{dA}{du} = -\frac{4}{5}\sqrt{5-u^2}$ . A chord of contact is drawn from the external point $P(4, 1)$ to the given ellipse. The area of the smaller region bounded by this chord of contact and the ellipse is:

Options

A.
$\pi - \frac{4}{5} - 2\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$
B.
$\frac{4}{5} + 2\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$
C.
$2\pi - \frac{4}{5} - 2\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$
D.
$\frac{\pi}{2} - \frac{4}{5} - \sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

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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

Standard Ellipse/Circle Area|Chord of Contact Logic|Integration Constant Trap

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to evaluate the area of an elliptical segment using a provided differential equation, combined with the geometric concept of the chord of contact.

The primary trap is the 'Integration Constant Trap'.

Students often integrate the given rate of change but forget to apply the boundary condition (the area is zero when the line becomes a tangent, i.e., $u=\sqrt{5}$ ).

Ignoring the constant $C$ leads to a negative area, which students might arbitrarily make positive, resulting in the incorrect option B.

This punishes rote integration without geometric grounding.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Standard Ellipse/Circle Area
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free