If P ( h , k ) is a point on the parabola x = 4 y 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 [2023]

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Solution

Q.
If $P (h, k)$ be a point on the parabola $x = 4 y^{2}$ , 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 [2023]

1 4

2 2

3 8

4 6

Ans.
(4)

We have $x = 4 y^{2}$

Equation of normal at $P (h, k)$

$y - k = - \frac{k}{(\frac{1}{8})} (x - h)$

$y - k = - 8 k (x - h)$

This line passes through the point $(0, 33)$ .

$33 - k = 8 k h$ ...(i)

$P (h, k)$ will satisfy $x = 4 y^{2}$ , so $h = 4 k^{2}$ . ...(ii)

Solving equations (i) and (ii), we get

$k = 1 and h = 4$

$∴$ $(h, k) \equiv (4, 1)$

Now we have equation of parabola, $y^{2} = 4 (x + y) \Rightarrow {(y - 2)}^{2} = 4 (x + 1)$

The directrix of this parabola is $x = - 2$

This is the line parallel to the y-axis.

So, distance of $P (4, 1)$ from the line $x = - 2$ is $6$ .

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