Question

A parabola, of latus rectum l, moves so as always to touch an equal parabola, their axes being parallel; prove that the locus of their point of contact is another parabola whose latus rectum is 2l.

Answer 1

Let $(h,k)$ be the co ordinates of the vertex of the moving parabola and its equation be ${(y-k)}^2+t(x-h)=0$. The equation of the fixed curve be $y^2-lx=0$ then the tangents at the point $(h^3,h)$ to the two curves are
$-2yt+x+l{t}^2=0$ and $y(lt-k)+\frac{1}{2}x+\frac{l^2{t}^2}{2}+k^2-lh-klt=0$
As they are co incident $\frac{k-lt}{2t}=\frac{l/2}{l}$
$=\frac{l^2{t}^2+2k^2-2lh-2klt}{2l{t}^2}$
From which ,
$t=\frac{k}{2l}$ and $k^2-lh=klt$
Eliminating, 't' we get $k^2=2lh$. Hence the locus of the vertex is $y^2=2lx$ which is a parabola having latus rectum as $2l$.