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Question #41566

Difficulty: 8/10 | 🟢 Free

The Question

Let $C_1$ be the curve $27x^2 = 16y^3$ and $C_2$ be the parabola $x^2 = 4y$ . Let $C_3$ be the envelope of the polars of points on $C_1$ with respect to $C_2$ . Find the area of the region bounded by $C_3$ , the line $3y + 1 = 0$ , and the vertical lines passing through the points of intersection of $C_2$ and the line $4y - 1 = 0$ .

Options

A.
$\frac{8}{3}$
B.
$\frac{4}{3}$
C.
$\frac{16}{3}$
D.
$\infty$

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

Parametric Curve Area|Intersection of Conics|Polar and Pole Duality|Bounded vs Unbounded Regions

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to identify improper integrals and unbounded regions disguised within a complex coordinate geometry setup.

By using Polar/Pole duality, the problem derives the envelope curve $y = -1/x^2$ .

The lethal trap lies in the final area calculation: students who blindly apply the Fundamental Theorem of Calculus across the singularity at $x = 0$ will evaluate $\int_{-1}^{1} (x^{-2} - 1/3) dx = [-x^{-1} - x/3]_{-1}^{1} = -8/3$ , and incorrectly conclude the area is $8/3$ (taking the absolute value).

However, because the integrand has an infinite discontinuity at the origin, the region is unbounded and the true area is infinite.

Quick Facts About This Question

Question Information

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