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Question #40508

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all points $P(x,y,z)$ in $\mathbb{R}^3$ such that the square of the distance from $P$ to the point $A(2, -1, 4)$ exceeds the square of the distance from $P$ to the $xy$ -plane by $12$ . Find the minimum possible distance from $P$ to the point $A$ .

Options

A.
$2\sqrt{3}$
B.
$\frac{7}{2}$
C.
$4$
D.
$2\sqrt{7}$

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

Maximum/Minimum of Quadratic Expressions|Locus via Distance Constraints|Distance Formula in 3D

Why This Question Is Tricky

Common Mistakes

This problem tests the student's ability to identify hidden domain constraints arising from 3D locus equations.

The trap lies in the relation $PA^2 = z^2 + 12$ .

A student might hastily minimize this by setting $z=0$ to get $PA = 2\sqrt{3}$ .

However, expanding the distance formula reveals that the locus is a paraboloid $(x-2)^2 + (y+1)^2 = 8z - 4$ .

The sum of squares on the LHS enforces the hidden quadratic constraint $8z - 4 \ge 0$ , restricting the domain to $z \ge 1/2$ .

This perfectly illustrates how geometric reality dictates algebraic bounds, a classic JEE Advanced pitfall where students linearize or unconstrain distance relationships.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Complex Numbers and Quadratic Equations - Class 11
Concept	Maximum/Minimum of Quadratic Expressions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free