Question

Distinct normals are drawn to the parabola y^2=4ax from the point (c 0)

Answer 1

Therefore we will get only one normal from $(1,0)$ or $(c,0)$

Step-by-step explanation:

Given;

$y^{2} =4ax$ and point $(c,0)$

Equation of normal of $m$ slope is,

$y=mx-2am-am^{3}$

For $y^{2} =4x$

$a=1$

Hence

$y=mx-2m-m^{3}$

Take $c=1$,

$(c,0)=(1,0)$  as it lies on normal,

$0=m-2m-m^{3}$

$m^{3} +2m-m=0$  ( multiplying with '-')

$m^{3} +m=0$

$m(m^{2}+1)=0$

$m=0$ or $m^{2} +1=0$

But $m^{2} +1=0$ is not possible for any real value of $x$,

hence only real solution is $m=0$

Hence we will get only one normal from $(1,0)$ or $(c,0)$