Question

Find the arithmetic mean of the co-ordinates of the feet of the normals from the point $(3,5)$ to the parabola $y^2=8x$.

Answer 1

Equation of parabola$=y^2=8x$

So, $4a=8$

$\Rightarrow a=2$

Equation of normal to parabola is

$yt+x-at(t^2+2)=0$

$\Rightarrow yt+x-2t(t^2+2)=0$

$\Rightarrow yt+x-2t^3-4t=0$

As normal posses through $(3,5)$

So,

$5t+3-2t^3-4t=0$

$\Rightarrow -2t^3=t+3=0$

It is a cubic equation in t.

Let $t,t_2,t_3$ be the roots

as coefficient of $t^2=0$

So $t_1+t_2+t_3=0$

$=4(t_1+t_2+t_3)=0$

$=4t_1+4t_2+4t_3=0$

$=y_1+y_2+y_3=0$

$\therefore$ Arithmetic mean$=0$.