Parabola
Nearest point on parabola, directrix distance
nta_pyq_2023_jan
Grade 11
Question:
If $P(h,k)$ be a point on the parabola $x = 4y^2$, 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:
Step-by-Step Solution
Key Concept: Minimize distance from point on $x=4y^2$ to Q using normal equation, find P, then compute distance to directrix of second parabola
On $x=4y^2$: parametrize as $(4t^2, t)$. Normal at $(4t^2, t)$ passes through $(0,33)$: $33 = t/8 + t^3/16 \Rightarrow t^3 + 2t - 528 = 0 \Rightarrow (t-8)(t^2+8t+66)=0 \Rightarrow t=8$. So $P=(4,1)$ (using $a=1/16$: $P=(1/16 \times 64, 2\times 1/16\times 8)=(4,1)$). For $y^2=4(x+y) \Rightarrow (y-2)^2 = 4(x+1)$: directrix is $x+1=-1 \Rightarrow x=-2$. Distance of $P(4,1)$ from $x=-2$ is $4-(-2)=6$. Answer: (4)
Correct Answer: 6