The parabolas `y^2=4ac` and `x^2=4by` intersect orthogonally at point `P(x_1,y_1)` where `x_1,y_1 != 0` provided (A) `b=a^2` (B) `b=a^3` (C) `b^3=a^2` (D) none of these

Options

A. `b = a^(2)`
B. `b = a^(3)`
C. `b^(3) = a^(2)`
D. None of these

Answer

D

Explanation

On solving `y^(2) = 4ax` and `x^(2) = 4by`, we get
`x =0` and `x^(3) = 64 ab^(2)`.
`((dy)/(dx))_(1)=((2a)/(y)),((dy)/(dx))_(n) =(x)/(2b)`
Given curves intersect orthogonally.
So, `(2a)/(y)xx (x)/(2b) =-1`
`rArr ax + by =0`
`rArr ax + (x^(2))/(4) =0` (From `x^(2) = 4by)`
`rArr x =- 4a (x ne0)`
`rArr -x^(3) = 64 a^(3)`
`rArr -64ab^(2) = 64 a^(3)`
`rArr a^(2)+b^(2) =0`
which is not possible