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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 :
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)\left( t ^2+8 t +66\right)=0 $
$ t =8$
$ P \left( at { }^2, 2 at \right)=\left(\frac{1}{16} \times 64,2 \times \frac{1}{16} \times 8\right)=(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)