If normal are drawn from a point `P(h , k)` to the parabola `y^2=4a x` , then the sum of the intercepts which the normals cut-off from the axis of the parabola is `(h+c)` (b) `3(h+a)` `2(h+a)` (d) none of these

Options

A. (h+a)
B. 3(h+a)
C. 2(h+a)
D. none of these

Answer

C

Explanation

(3) The equation of normal is `y=mx-2am-am^(3)`.
Putting y=0, we get
`x_(1)=2a+am_(1)^(2)`
`x_(2)=2a+am_(2)^(2)`
`andx_(3)=2a+am_(3)^(2)`
where `x_(1),x_(2)andx_(3)` are the intercepts on the axis of the parabola.
The normal passes through (h,k). Therefore,
`am^(3)+(2a-h)m+k=0`
It has roots `m_(1),m_(2)andm_(3)` which are the slopes of the normals.
Therefore,
`m_(1)+m_(2)+x_(3)=0`
`andm_(1)m_(2)+m_(2)m_(3)+m_(3)m_(1)=(2a-h)/(a)`
`:.m_(1)^(2)+m_(2)^(2)+m_(3)^(2)=(m_(1)+m_(2)+m_(3))^(2)-2(m_(1)m_(2)+m_(2)m_(3)+m_(3)m_(1))`
`=-(2(2a-h))/(a)`
`orx_(1)+x_(2)+x_(3)=6a-2(2a-h)=2(h+a)`