Question

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 :

• A. 4
• B. 2
• C. 8
• D. 6

Answer 1

Answer: (D)

6

Answer 2

Equation of normal
y=−tx+2at+at3
y=−tx+162​t+161​t3
It passes through (0,33)
33=8t​+16t3​
t3+2t−528=0
(t−8)(t2+8t+66)=0
t=8
P(at2,2at)=(161​×64,2×161​×8)=(4,1)
Parabola :
y2=4(x+y)
⇒y2−4y=4x
⇒(y−2)2=4(x+1)
Equation of directix :-
x+1=−1
x=−2
Distance of point =6
Ans. : (4)