If `x+y=k` is normal to `y^2=12 x ,` then `k` is (a)3 (b) 9 (c) `-9` (d) `-3`

Options

A. 3
B. 9
C. -9
D. -3

Answer

B

Explanation

We have,
`y^(2)=12x rArr 2y(dy)/(dx) =12 rArr (dy)/(dx) =(6)/(y) `
Let ` x+y=k ` be normal to ` y^(2)=12x ` at point P`(x_(1),y_(1)). ` Then,
` (-(1)/((dy)/(dx)))_("at" P )`= (Slope of the line `x+y=k` )
` rArr -(y_(1))/(6) =-1 rArr y_(1)=6 `
Since `(x_(1),y_(1))` lies on ` y^(2)=12x ` Therefore ,
`y_(1)^(2)=12x_(1) rArr 12_(1) =36 rArr x_(1)=3 `
Also, `P(x_(1),y_(1)) ` lies on ` x+y=k. ` Therefore,
` x_(1)+y_(1)=k rArr k=9 `