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

Let $A$ be a point on the $x$ -axis. Common tangents are drawn from $A$ to the curves $x^2 + y^2 = 8$ and $y^2 = 16x$ . If one of these tangents touches the two curves at $Q$ and $R$ , then $(QR)^2$ is equal to:

Choose the correct answer:

Explanation

Solution

Parabola $y^2 = 16x$ ki tangent ki equation: $y = mx + \frac{4}{m}$ .

Circle $x^2 + y^2 = 8$ ke liye tangent ki condition: $c^2 = r^2(1 + m^2)$ .

\left( \frac{4}{m} \right)^2 = 8(1 + m^2) \implies \frac{16}{m^2} = 8(1 + m^2)

2 = m^2 + m^4 \implies m^4 + m^2 - 2 = 0 \implies (m^2 + 2)(m^2 - 1) = 0

Chunki $m$ real hai, $m^2 = 1 \implies m = \pm 1$ .

Points of contact nikalne par:

Parabola par $R = (\frac{a}{m^2}, \frac{2a}{m}) = (4, 8)$ (agar $m=1$ lein).
Circle par $Q = (\frac{-r^2 m}{c}, \frac{r^2}{c}) = (-2, 2)$ .

(QR)^2 = (4 - (-2))^2 + (8 - 2)^2 = 6^2 + 6^2 = 36 + 36 = 72

Correct Option: (2)