### Parabola Question 4

#### Question 4 - 30 January - Shift 1

If $P(h, k)$ be 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 :

(1) 2

(2) 4

(3) 8

(4) 6

## Show Answer

#### Answer: (4)

#### Solution:

#### Formula: Equation of normal at any Point in Parametric form, Equation of Directrix

Equation of normal

$y=-t x+2 a t+a t^{3}$

$y=-t x+\frac{2}{16} t+\frac{1}{16} t^{3}$

It passes through $(0,33)$

$33=\frac{t}{8}+\frac{t^{3}}{16}$

$t^{3}+2 t-528=0$

$(t-8)(t^{2}+8 t+66)=0$

$t=8$

$P(at^{2}, 2 at)=(\frac{1}{16} \times 64,2 \times \frac{1}{16} \times 8)=(4,1)$

Parabola :

$y^{2}=4(x+y)$

$\Rightarrow y^{2}-4 y=4 x$

$\Rightarrow(y-2)^{2}=4(x+1)$

Equation of directix :-

$x+1=-1$

$x=-2$

Distance of point $=6$

Ans. : (4)