JEE 2023 — Mathematics PYQ

Q: How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The line $x=8$ is the directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with focus $(2, 0)$ . If the tangent to $E$ at point $P$ in the first quadrant passes through $(0, 4\sqrt{3})$ and intersects the $x$ -axis at $Q$ , then $(3PQ)^2$ is equal to:

Choose the correct answer:

Explanation

Solution:

Ellipse Properties: Focus $ae = 2$ , Directrix $a/e = 8$ .

Multiplying: $a^2 = 16 \implies a = 4$ .

$e^2 = 2/8 = 1/4 \implies b^2 = a^2(1-e^2) = 16(3/4) = 12$ .

Ellipse: $\frac{x^2}{16} + \frac{y^2}{12} = 1$

Tangent: Tangent at $(x_1, y_1)$ passes through $(0, 4\sqrt{3})$ .

$\frac{x \cdot 0}{16} + \frac{y \cdot 4\sqrt{3}}{12} = 1 \implies y_1 = \sqrt{3}$ .

Since $P$ is on ellipse: $\frac{x_1^2}{16} + \frac{3}{12} = 1 \implies x_1^2 = 12 \implies x_1 = 2\sqrt{3}$ .

$P = (2\sqrt{3}, \sqrt{3})$ .

Point $Q$ : Tangent at $P$ : $\frac{x(2\sqrt{3})}{16} + \frac{y(\sqrt{3})}{12} = 1$ .

For $x$ -intercept $Q$ , set $y=0$ : $x_Q = \frac{16}{2\sqrt{3}} = \frac{8}{\sqrt{3}}$ .

$Q = (8/\sqrt{3}, 0)$ .

Length $PQ$ : $PQ^2 = (2\sqrt{3} - 8/\sqrt{3})^2 + (\sqrt{3}-0)^2 = (-2/\sqrt{3})^2 + 3 = 4/3 + 3 = 13/3$ .

Final Value: $(3PQ)^2 = 9 \times \frac{13}{3} = 39$ .