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Question #40425

Difficulty: 8/10 | 🟢 Free

The Question

Consider the ellipse $E: \frac{(x-1)^2}{9} + \frac{(y-2)^2}{25} = 1$ . A tangent drawn from the point $P\left(-2, \frac{9}{2}\right)$ to $E$ intersects the upper latus rectum of $E$ at a point $Q$ . The sum of all possible values of the abscissa of $Q$ is:

Options

A.
$-\frac{14}{5}$
B.
$-\frac{4}{5}$
C.
$-2$
D.
$-\frac{24}{5}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Tangent Half-Angle Substitution|Latus Rectum Geometry|Slope Sensitivity

Why This Question Is Tricky

Common Mistakes

This question tests the 'Slope Sensitivity' trap associated with the Weierstrass substitution $t = \tan(\theta/2)$ in coordinate geometry.

When finding tangents from an external point, students often convert the trigonometric tangent equation into a polynomial in $t$ .

If the external point lies on a vertical tangent, the leading coefficient of the $t$ -polynomial vanishes, corresponding to the root $t \to \infty$ (or $\theta = \pi$ ).

Students who mechanically solve for $t$ will only find the finite root, missing the vertical tangent entirely, leading to an incomplete set of intersection points with the latus rectum.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Tangent Half-Angle Substitution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free