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Question #41571

Difficulty: 8/10 | 🟢 Free

The Question

Let $S_1$ be the area of the finite region bounded by the parabola $y^2 = 4x$ , its latus rectum, and the tangent to the parabola at the point $P(t^2, 2t)$ , where $t > 1$ . Let $S_2$ be the area of the region bounded by the parabola, its latus rectum, and the x-axis in the first quadrant. If $S_1 = S_2$ , then the value of $t^2$ is

Options

A.
$3 + 2\sqrt{3}$
B.
$4$
C.
$9$
D.
$3 + \sqrt{3}$

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

Region Ambiguity|Latus Rectum Geometry|Condition of Tangency

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly identify dynamic integration bounds based on a parameter (Region Ambiguity).

The lethal trap lies in assuming the region $S_1$ starts from $x=0$ (yielding $t^2=4$ ) instead of the latus rectum $x=1$ .

Because the tangent parameter $t > 1$ , the point of tangency $x=t^2$ is to the right of the latus rectum, forcing the integration interval to be $[1, t^2]$ .

A secondary trap involves misreading $S_2$ as the entire area bounded by the latus rectum across both quadrants, which leads to the perfectly factorable incorrect equation $t^4 - 6t^2 - 8t - 3 = 0$ and the trap answer $t^2=9$ .

The clean integer trap answers heavily penalize students who do not rigorously sketch the region.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Region Ambiguity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free