JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $P(h, k)$ be a point on the parabola $x = 4y^2$ , which is nearest to the point $Q(0, 33)$ , then the distance of $P$ from the directrix of the parabola $y^2 = 4(x + y)$ is equal to:

Choose the correct answer:

Explanation

Solution

Parabola $x = 4y^2 \implies y^2 = \frac{1}{4}x$ . Point $P$ is $(4k^2, k)$ .

Distance $PQ$ minimize karne ke liye normal $P$ se $Q(0, 33)$ pass honi chahiye.

Normal equation at $k$ : $y - k = -8k(x - 4k^2)$ .

Pass through $(0, 33)$ : $33 - k = -8k(0 - 4k^2) \implies 33 - k = 32k^3 \implies 32k^3 + k - 33 = 0$

Solving gives $k = 1$ . So, $P = (4, 1)$ .

Second parabola: $y^2 - 4y = 4x \implies (y - 2)^2 = 4(x + 1)$ .

Iski directrix: $X = -A \implies x + 1 = -1 \implies x = -2$ .

Distance of $P(4, 1)$ from $x + 2 = 0$ is $|4 + 2| = 6$ .

Correct Option: (2)

Choose the correct answer:

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