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Question #40824

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \int_0^x (\sqrt{3} + \cos t) dt$ . By the Lagrange Mean Value Theorem, there exists at least one point $c \in (0, 2\pi)$ where the tangent line $T$ to the curve $y = f(x)$ is parallel to the secant line joining the points $(0, f(0))$ and $(2\pi, f(2\pi))$ . A point $P$ lies on the ellipse $E: x^2 + 3y^2 = 12$ such that the normal to $E$ at $P$ is parallel to $T$ . If $P$ lies on the arc of the ellipse where $y$ is a strictly convex function of $x$ , find the distance of $P$ from the focus of $E$ having a positive $x$ -coordinate.

Options

A.
$2\sqrt{3} + 2$
B.
$2\sqrt{3} - 2$
C.
$2\sqrt{3} + \sqrt{2}$
D.
$2\sqrt{3} - \sqrt{2}$

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

Normal to Conic|LMVT Tangent Parallelism|Concavity Analysis|Symmetry and Periodicity

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of calculus (LMVT and concavity) with coordinate geometry (normals to conics).

The primary trap lies in the symmetry of the parametric form: solving for the normal's slope yields two diametrically opposite points on the ellipse ( $\theta = \pi/4$ and $5\pi/4$ ).

Students often blindly pick the first quadrant angle.

However, the geometric constraint that the curve must be strictly convex restricts the valid domain to the lower half of the ellipse ( $y < 0$ ), forcing $\theta = 5\pi/4$ .

Failing to apply this concavity analysis leads to calculating the distance from the wrong point, resulting in the trap option $2\sqrt{3} - 2$ .

Quick Facts About This Question

Question Information

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