Question

Prove that the locus of the vertices of all parabolas that can be drawn touching a given circle of radius a and having a fixed point on the circumference as focus is $r=2a{\cos }^3\frac{\theta }{3},$ the fixed point being the pole and the diameter through it the initial line.

Answer 1

If the origin lies on the circumference of the circle, and taking the initial line as the diameter, its equation may be given by
$r=2a\cos \theta$ .... $\left(1\right)$
Hence taking the fixed point as origin, let the circle be represented by $\left(1\right)$.

If P be any point such that for $P,\theta =\alpha$, then the equation to the tangent at P may be given by
$r\cos (\theta -2\alpha )=2a{\cos }^2\alpha$ ..... $\left(2\right)$
If it meets the axis of the parabola at T, then we have $SP=ST$ .... $\left(3\right)$
If $\beta$ be the vectorial angle of T then putting in $2$, we get
$r\cos (\beta -2\alpha )=2a{\cos }^2\alpha$
$r=\frac{2a{\cos }^2\alpha }{\cos (\beta -2\alpha )}=ST$ .... $\left(4\right)$
As the vectorial angle of P is $2\alpha$ and it lies on the circle, hence by $\left(1\right)$
$SP=2a\cos \alpha$ ..... $\left(5\right)$
By $\left(3\right),\left(4\right)$ & $\left(5\right)$ we get
$2a\cos \alpha =\frac{2a{\cos }^2\alpha }{\cos (\beta -2\alpha )}$
or $\cos (\beta -2\alpha )=\cos \alpha$
$\beta -2\alpha =\alpha$
$\beta =3\alpha$ ..... $\left(6\right)$
In the same figure, we have
$SN=SP\cos (\beta -\alpha )=2a\cos \alpha \cos 2\alpha$
$\therefore 2SA=ST+SN=2a\cos \alpha +2a\cos \alpha \cos 2\alpha$
$=2a\cos \alpha (1+\cos 2\alpha )$
$=4a{\cos }^3\alpha$
$=4a{\cos }^3\frac{\beta }{3}$
or $AS=2a{\cos }^3\frac{\beta }{3}$
$\therefore$ Locus of A is $r=2a{\cos }^3\frac{\theta }{3}$