Let PQ be a focal chord of the parabola y 2 = 36 x of length 100, making an acute angle with the positive x -axis. Let the ordinate of P be positive and M be the point on the line segment PQ such that PM : MQ = 3 : 1. Then which of the following points

Question

Let PQ be a focal chord of the parabola $y^{2} = 36 x$ of length 100, making an acute angle with the positive $x$ -axis. Let the ordinate of P be positive and M be the point on the line segment PQ such that PM : MQ = 3 : 1. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line PQ? [2023]

Smart Achievers · Accepted Answer

(4)

$We have y^{2} = 36 x \dots (i)$

$Length of P Q = 100 units$

$P M : M Q = 3 : 1$

$y^{2} = 4 a x \dots (i i)$

$On comparing (i) and (ii), we get a = 9$

$∴$ $Now, the coordinates of P \equiv (9 t^{2}, 18 t)$ $and coordinate of Q \equiv (\frac{9}{t^{2}}, \frac{18}{t})$

$P Q = a {(t + \frac{1}{t})}^{2} \Rightarrow 100 = {(t + \frac{1}{t})}^{2}$

$\Rightarrow t + \frac{1}{t} = \pm \frac{10}{3}$

$\Rightarrow 3 t^{2} + 3 = 10 t \Rightarrow 3 t^{2} - 10 t + 3 = 0 \Rightarrow (3 t - 1) (t - 3) = 0$

$\Rightarrow t = \frac{1}{3}, 3$

or $t + \frac{1}{t} = \frac{- 10}{3} \Rightarrow 3 t^{2} + 3 = - 10 t \Rightarrow 3 t^{2} + 10 t + 3 = 0$

$\Rightarrow (3 t + 1) (t + 3) = 0 \Rightarrow t = \frac{- 1}{3}, - 3$

$∴$ $Coordinates of P are (81, 54) and coordinates of Q are (1, - 6)$

$Now, the coordinates of M are$ $(\frac{3 (1) + 81}{4}, \frac{3 (- 6) + 54}{4}) \equiv (21, 9)$

$Slope of P Q = \frac{- 6 - 54}{1 - 81} = \frac{3}{4}$

$∴$ $Slope of the line perpendicular to P Q = \frac{- 4}{3}$

$Therefore, the equation of line passing through M and perpendicular to P Q is given by$

$y - 9 = \frac{- 4}{3} (x - 21) \Rightarrow 4 x + 3 y = 111$

$Only point (- 3, 43) does not satisfy the equation 4 x + 3 y = 111 .$

Solution

1 (3, 33)

2 (- 6, 45)

3 (6, 29)

4 (- 3, 43)

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