Question

Two equal parabolas have the same vertex and their axes are at right angles ; prove that the common tangent touches each at the end of a latus rectum.

Answer 1

Let the two parabolas be $y^2=4ax......\left(i\right)$ and $x^2=4ay......\left(ii\right)$

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

$y=mx+\frac{a}{m}$

It also touches $\left(ii\right)$ so substituting $y$ in $\left(ii\right)$

$x^2=4a(mx+\frac{a}{m})m{x}^2=4a{m}^{2}x+4a^{2}m{x}^2-4a{m}^{2}x-4a^2=0$

This is quadratic in $x$ and will have only one root

$\Rightarrow {(-4a{m}^2)}^2-4\left(m\right)(-4a^2)=016a^2{m}^4+16a^{2}m=0\Rightarrow m=-1$

Point of contact to $\left(i\right)$ is $(\frac{a}{m^2},\frac{2a}{m})$

$\Rightarrow (a,-2a)$

Point of contact to $\left(ii\right)$ is $(\frac{2a}{m},\frac{a}{m^2})$

$\Rightarrow (-2a.a)$

which are respective ends of their latusrectum

Hence proved.