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 ( P Q ) 2 is [2023]

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Solution

Q.
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 ${(P Q)}^{2}$ is [2023]

1 296

2 325

3 346

4 317

Ans.
(2)

$We have y^{2} = 20 x$    $\dots (i)$

$∴$ $Focus, R \equiv (5, 0)$

$Now, put x = \frac{y - c}{m} in y^{2} = 20 x$

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

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

$\Rightarrow m y^{2} - 20 y + 20 (6 + m) = 0$    $(∵ c - m = 6 given)$

$Let y_{1}$ and $y_{2} be the roots, then$

$y_{1} + y_{2} = \frac{20}{m}$ $\dots (i i)$

Now centroid of $∆ P Q R$ is at (10,10)

$∴$ $\frac{y_{1} + y_{2} + 0}{3} = 10$

$\Rightarrow y_{1} + y_{2} = 30$    $\dots (i i i)$

$From (ii) and (iii), we get$

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

$Now, c = 6 + m = 6 + \frac{2}{3} = \frac{20}{3}$

$∴$ $Equation of line P Q is y = \frac{2}{3} x + \frac{20}{3}$

$\Rightarrow 2 x - 3 y + 20 = 0$    $\dots (i v)$

$Now, solving equation (i) and (iv), we get$

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

$∴$    ${(P Q)}^{2} = 15^{2} + 10^{2} = 325$

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