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Question #40891

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be a surface defined by the locus of points $P(x,y,z)$ satisfying the equation $d(P, F_1) + d(P, F_2) + 2z \cdot \text{sgn}(z^2) = 4$ where $F_1(1, 0, 0)$ and $F_2(-1, 0, 0)$ are fixed points, $d(A,B)$ is the Euclidean distance between $A$ and $B$ , and $\text{sgn}(\cdot)$ denotes the signum function. Let $L$ be the normal line to the surface $S$ at the point $P_0(0, \sqrt{3}, 0)$ . If $L$ intersects the $xz$ -plane at the point $Q$ , what is the distance $P_0Q$ ?

Options

A.
$\sqrt{7}$
B.
$\sqrt{3}$
C.
$2\sqrt{2}$
D.
$3$

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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

Distance Formula in 3D|Tangent/Normal Slope Interplay|Signum Function Discontinuity

Why This Question Is Tricky

Common Mistakes

This question tests the interplay between 3D geometry (tangent/normal planes of an ellipsoid-like surface) and fundamental calculus concepts.

The core trap lies in the term $2z \text{sgn}(z^2)$ .

Students often blindly apply the product rule, yielding $2\text{sgn}(z^2) + 2z \cdot \frac{d}{dz}\text{sgn}(z^2)$ .

At $z=0$ , they incorrectly evaluate this as $2(0) + 0 = 0$ , leading to a flawed normal vector $\langle 0, \sqrt{3}, 0 \rangle$ and an incorrect distance of $\sqrt{3}$ .

The rigorous approach requires recognizing that $z \text{sgn}(z^2)$ simplifies to $z$ for all $z \neq 0$ , meaning its derivative at $z=0$ must be evaluated using first principles (limits), which correctly yields 2.

This forces a limit-based evaluation of the gradient, penalizing formulaic algebraic substitution.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Distance Formula in 3D
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free