Question

$ABCD$ is a rectangular sheet of paper, and it is folded over so that $C$ lies on the side $AB$; prove that the envelope of the crease so formed is a parabola, whose focus is the initial position of $C$.

Answer 1

Let $CB$ and $CD$ be the axes, and let $CB=a$
The crease will be a straight line.

Let its equation be $x\cos \theta +y\sin \theta =p$. So we get
$a=2p\cos \theta$
Now eliminating '$p$' the equation becomes
$2x{\cos }^2\theta +2y\sin \theta \cos \theta =a({\cos }^2\theta +{\sin }^2\theta )$
or $(2x-a){\cos }^2\theta +2y\sin \theta \cos \theta -a{\sin }^2\theta =0$
Equating its discriminant equal to zero, the required envelope is
$4y^2{\sin }^2\theta -4\left[(2x-a)(-a{\sin }^2\theta )\right]=0$
$y^2+a(2x-a)=0$
This is a parabola whose focus is at $C$.