Question

Parabolas are drawn to touch two given straight lines which are inclined at an angle $\omega$; if the chords of contact all pass through a fixed point, prove that
(1) their directrices all pass through another fixed point, and (2) their foci all lie on a circle which goes through the intersection of the two given straight lines.

Answer 1

Equation of the directrix is given by

$\frac{x+y\cos \omega }{a}+\frac{y+x\cos \omega }{b}=\mathrm{1..........}\left(i\right)$

As the chord passes through fixed point whose coordinates are $(h,k)$ we have

$\frac{h}{a}+\frac{k}{b}=\mathrm{1........}\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$

$x+y\cos \omega =h,y+x\cos \omega =k$

Solving the equations

$\Rightarrow x=\frac{h-k\cos \omega }{1-{\cos }^2\omega },y=\frac{k-h\cos \omega }{1-{\cos }^2\omega }$

So, the directrix passes through fixed point $(\frac{h-k\cos \omega }{1-{\cos }^2\omega },\frac{k-h\cos \omega }{1-{\cos }^2\omega })$

Hence proved.

(2) Focus is given by

$ax=by=x^2+{2xy\cos \omega +y}^2$

$\Rightarrow a=\frac{x^2+{2xy\cos \omega +y}^2}{x},b=\frac{x^2+{2xy\cos \omega +y}^2}{y}$

Substituting $a$ and $b$ in $\left(ii\right)$

$\frac{hx}{x^2+{2xy\cos \omega +y}^2}+\frac{ky}{x^2+{2xy\cos \omega +y}^2}=1hx+ky=x^2+{2xy\cos \omega +y}^2{x}^2+{2xy\cos \omega +y}^2-hx-ky=0$

Clearly the equation represents a circle.