Question

If a parabola, whose latus rectum is 4c, slide between two rectangular axes, prove that the locus of its focus is $x^2{y}^2=c^2(x^2+y^2)$, and that the curve traced out by its vertex is $x^{\frac{2}{3}}{y}^{\frac{2}{3}}(x^{\frac{2}{3}}+y^{\frac{2}{3}})=c^2$.

Answer 1

When axis is at some angle $\omega$ then the focus lies on intersection of

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

Here $\omega =\frac{\pi }{2}$

$\Rightarrow x^2+y^2=axa=\frac{x^2+y^2}{x}........\left(i\right)$

Also $x^2+y^2=by$

$\Rightarrow b=\frac{x^2+y^2}{y}.......\left(ii\right)$

Now $c^2{(a^2+b^2)}^3=a^4{b}^4.....\left(iii\right)$

Substituting $\left(i\right)$ and $\left(ii\right)$, we get

$c^2{\{{\left(\frac{x^2+y^2}{x}\right)}^2+{\left(\frac{x^2+y^2}{y}\right)}^2\}}^3={\left(\frac{x^2+y^2}{x}\right)}^4{\left(\frac{x^2+y^2}{y}\right)}^4{c}^2{(x^2+y^2)}^6{\{{\left(\frac{1}{x}\right)}^2+{\left(\frac{1}{y}\right)}^2\}}^3=\frac{{(x^2+y^2)}^8}{x^4{y}^4}{c}^2{(x^2+y^2)}^6\frac{{(x^2+y^2)}^3}{x^6{y}^6}=\frac{{(x^2+y^2)}^8}{x^4{y}^4}{c}^2(x^2+y^2)=x^2{y}^2$

Hence proved

Coordinates of vertex when parabola is inclined are

$(\frac{a{b}^2{(b+a\cos \omega )}^2}{{(a^2+2ab\cos \omega +b^2)}^2},\frac{b{a}^2{(a+b\cos \omega )}^2}{{(a^2+2ab\cos \omega +b^2)}^2})$

Here $\omega =90$

$\Rightarrow x=\frac{a{b}^4}{{(a^2+b^2)}^2},y=\frac{b{a}^4}{{(a^2+b^2)}^2}\frac{x}{y}=\frac{b^3}{a^3}\Rightarrow \frac{b}{x^{\frac{1}{3}}}=\frac{a}{y^{\frac{1}{3}}}=\alpha \left(suppose\right)..........\left(iv\right)$

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

$c^2=\frac{{\alpha }^3{x}^{\frac{4}{3}}{y}^{\frac{4}{3}}}{x^{\frac{2}{3}}+x^{\frac{2}{3}}}$

From $\left(iii\right)$

$1=\frac{\alpha x^{\frac{1}{3}}{y}^{\frac{1}{3}}}{(x^{\frac{2}{3}}+x^{\frac{2}{3}})}$

Eliminating $\alpha$

$x^{\frac{2}{3}}{y}^{\frac{2}{3}}(x^{\frac{2}{3}}+x^{\frac{2}{3}})=c^2$

Hence proved.