Question

A parabola is drawn to pass through $A$ and $P$, the ends of a diameter of a given circle of radius a, and to have as directrix a tangent to a concentric circle of radius $h$ ; the axes being AB and a perpendicular diameter, prove that the locus of the focus of the parabola is $\frac{x^2}{b^2}+\frac{y^2}{b^2-a^2}=1.$

Answer 1

Let the equation of circle be $x^2+y^2=a^2$ with ends of diameter $M(a,0)$ and $M^{\prime }(-a,0)$

Equation of concentric circle is $x^2+y^2=b^2........\left(i\right)$

Equation of tangent to $\left(i\right)$ is

$y=mx+b\sqrt{m^2+1}$

$mx-y+b\sqrt{m^2+1}=0$

which is the directrix of parabola

Let the focus be $S(h,k)$

For parabola we have $PS=PM$

$\sqrt{{(h-a)}^2+k^2}=\frac{m\left(a\right)-1\left(0\right)+b\sqrt{m^2+1}}{\sqrt{m^2+1}}\sqrt{{(h-a)}^2+k^2}=\frac{am+b\sqrt{m^2+1}}{\sqrt{m^2+1}}......\left(ii\right)$

Also $PS=P{M}^{\prime }$

$\sqrt{{(h+a)}^2+k^2}=\frac{m(-a)-1\left(0\right)+b\sqrt{m^2+1}}{\sqrt{m^2+1}}\sqrt{{(h+a)}^2+k^2}=\frac{-am+b\sqrt{m^2+1}}{\sqrt{m^2+1}}.....\left(iii\right)$

Adding $\left(ii\right)$ and $\left(iii\right)$

$\sqrt{{(h-a)}^2+k^2}+\sqrt{{(h+a)}^2+k^2}=2b\sqrt{{(h+a)}^2+k^2}=2b-\sqrt{{(h-a)}^2+k^2}$

squaring both sides

$h^2+2ah+a^2+k^2={4b}^2+h^2-2ah+a^2+k^2-4b\sqrt{{(h-a)}^2+k^2}{4b}^2-4ah=4b\sqrt{{(h-a)}^2+k^2}{b}^2-ah=b\sqrt{{(h-a)}^2+k^2}b-\frac{ah}{b}=\sqrt{{(h-a)}^2+k^2}$

Again squaring both sides

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

generalising the equation

$\frac{x^2}{b^2}+\frac{y^2}{b^2-a^2}=1$