JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let R be the focus of the parabola $y^{2} = 20 x$ and the line $y = m x + c$ intersect the parabola at two points P and Q.

Let the point G(10, 10) be the centroid of the triangle PQR.If $c - m = 6$ ,then $(PQ)^{2}$ is

Choose the correct answer:

Explanation

$y^{2} = 20 x, y = m x + c$

Put value of $x$

$y^{2} = 20 (\frac{y - c}{m})$

$\Rightarrow y^{2} - \frac{20}{m} y + \frac{20}{m} c = 0$

Since, centroid = (10, 10)

So,

$\frac{y _{1} + y _{2} + 0}{3} = 10$

$\Rightarrow y_{1} + y_{2} = 30$

From (1),

Sum of roots = $\frac{20}{m} = 30 \Rightarrow m = \frac{2}{3}$

Also, $c - m = 6 \Rightarrow c = 6 + \frac{2}{3} = \frac{20}{3}$

Now, the equation is:

$y^{2} - \frac{20}{3} \times 3 y + \frac{20}{2} \times \frac{20}{3} = 0$

$\Rightarrow y^{2} - 30 y + 200 = 0$

$\Rightarrow y^{2} - 20 y - 10 y + 200 = 0$

$\Rightarrow (y - 20) (y - 10) = 0$

$\Rightarrow y = 10, 20 \Rightarrow x = 5, 20$

$∴ P \equiv (5, 10), Q \equiv (20, 20)$

So,

$P Q^{2} = (20 - 5)^{2} + (20 - 10)^{2}$

$= 225 + 100 = 325$