Question

Prove that the locus of the middle point of the portion of a normal intersected between the curve and the axis is a parabola whose vertex is the focus and whose latus rectum is one quarter of that of the original parabola.

Answer 1

Let the equation of parabola be $y^2=4ax$ and the middle point of intercepted portion be $(h,k)$

Equation of normal at $P(a{t}^2,2at)$ is

$y=-tx+2at+a{t}^3$

Put $y=0$

$0=-tx+2at+a{t}^3\Rightarrow x=2a+a{t}^2$

So, the point of intersection with axis is $Q(2a+a{t}^2,0)$

Mid point of $PQ$ is

$(\frac{a{t}^2+2a+a{t}^2}{2},\frac{2at+0}{2})\left(a\right(1+t^2),at)a(1+t^2)=h....\left(i\right)at=k\Rightarrow t=\frac{k}{a}$

Substitute $t$ in $\left(i\right)$

$a(1+{\left(\frac{k}{a}\right)}^2)=h{k}^2+a^2=ah{k}^2=a(h-a)$

So, the general equation is

$y^2=a(x-a)$

Length of latus rectum $=4\times \frac{a}{4}=a$

which is one quarter the latusrectum of orignal parabola

Vertex of parabola is

$x-a=0,y=0\Rightarrow (a,0)$

which is focus of orignal parabola

Hence proved.