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Question #40826

Difficulty: 8/10 | 🟢 Free

The Question

Let $H$ be the hyperbola $\frac{x^2}{36} - \frac{y^2}{4} = 1$ . For any point $P(x, y)$ on $H$ with $y > 0$ , let $A(x)$ be the area of the region bounded by $H$ , the x-axis, and the normal to $H$ at $P$ . Let $L = \lim_{x \to -\infty} x^2 \left( A(x) - \frac{5}{27} x^2 + 6 \ln|x| + \frac{10}{3} - 6 \ln 3 \right)$ Find the value of $6L + 5$ .

Options

A.
149
B.
18
C.
-139
D.
125

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|Parametric Curve Area|Indeterminate Forms|Real-Valued Function Domain

Why This Question Is Tricky

Common Mistakes

This question tests the asymptotic behavior of the area bounded by a hyperbola and its normal, specifically targeting the 'Real-Valued Function Domain' and 'Indeterminate Forms' traps.

Students frequently memorize the indefinite integral of $\sqrt{x^2-a^2}$ without the absolute value in the logarithmic term.

When evaluating the area for the left branch ( $x \to -\infty$ ), omitting the absolute value results in a negative argument for the logarithm, leading students to incorrectly conclude the limit does not exist or to make forced algebraic errors.

Properly applying the absolute value reveals the area function is even, allowing a standard binomial expansion to resolve the $O(1/x^2)$ indeterminate form, yielding the precise calculated trap value.

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