Question

A parabola touches two given straight lines; if its axis pass through the point (h, k), the given lines being the axes of coordinates, prove that the locus of the focus is the curve
$x^2-y^2-hx+ky=0$.

Answer 1

When axes is inclined focus is given by

$ax=by=x^2+{2xy\cos \omega +y}^2\Rightarrow a=\frac{x^2{+y}^2}{x},b=\frac{x^2{+y}^2}{y}.......\left(i\right)$

The abssica of focus is $x=\frac{a{b}^2}{a^2+{2ab\cos \omega +b}^2}..........\left(ii\right)$

Axis of parabola is given by

$ay-bx=\frac{ab(a^2-b^2)}{a^2+{2ab\cos \omega +b}^2}$

It passes through $(h,k)$

$ak-bh=\frac{ab(a^2-b^2)}{a^2+{2ab\cos \omega +b}^2}ak-bh=\frac{a{b}^2}{a^2+{2ab\cos \omega +b}^2}\times \frac{a^2-b^2}{b}$

Substituting $\left(ii\right)$

$ak-bh=\frac{{x(a}^2-b^2)}{b}$

Substituting $a$ and $b$ from $\left(i\right)$

$\left(\frac{x^2{+y}^2}{x}\right)k-\left(\frac{x^2{+y}^2}{y}\right)h=\frac{x\{{\left(\frac{x^2{+y}^2}{x}\right)}^2-{\left(\frac{x^2{+y}^2}{y}\right)}^2\}}{\left(\frac{x^2{+y}^2}{y}\right)}\frac{k}{x}-\frac{h}{y}=xy(\frac{1}{x^2}-\frac{1}{y^2})x^2-y^2-hx+ky=0$

Hence proved.