Question

The ordinates of the foot of normal(s) to the parabola $y^2=4ax$ from the point $(6a,0)$ is/are

• A. $0$
• B. $4a$
• C. $-2a$
• D. $2a$

Answer 1

Answer: (A), (B)

$4a$

Normal to $y^2=4ax$ is
$y+tx=a{t}^3+2at$
Since $(6a,0)$ lies on it, so
$0+6at=a{t}^3+2at\Rightarrow 6t=t^3+2t(\because a\ne 0)\Rightarrow t(t-2)(t+2)=0\therefore t=-2,0,2$

Coordinates of foot of normal are
$(a{t}_1^2,2a{t}_1),(a{t}_2^2,2a{t}_2),(a{t}_3^2,2a{t}_3)\Rightarrow (4a,-4a),(0,0),(4a,4a)$

$\therefore$ Ordinates are $-4a,0,4a$