Question

If $a^2>8b^2,$ prove that a point can be found such that the two tangents from it to the parabola $y^2=4ax$ are normals to the parabola $x^2=4by.$

Answer 1

Equation of tangent to $y^2=4ax$ is

$t^{\prime }y=x+a{t^{\prime }}^{2}x-t^{\prime }y+a{t^{\prime }}^2=\mathrm{0......}\left(i\right)$

Parametric point for $x^2=4by$ is $P(2bt,b{t}^2)$

$2x=4b\frac{dy}{dx}\frac{dy}{dx}=\frac{x}{2b}$

$\therefore$ Slope of tangent at $P$ is

$m=\frac{2bt}{2b}=t$

$\Rightarrow$slope of normal $=-\frac{1}{t}$

Equation of normal at $P$ is

$y-b{t}^2=-\frac{1}{t}(x-2bt)ty-b{t}^3=-x+2btx+ty-b{t}^3-2bt=\mathrm{0........}\left(ii\right)$

Now the tangent and normal are common, so $\left(i\right)$ and $\left(ii\right)$ are equation of same line

$\frac{1}{1}=\frac{-t^{\prime }}{t}=\frac{a{t^{\prime }}^2}{-b{t}^3-2bt}\Rightarrow t^{\prime }=-t-b{t}^3-2bt=a{t^{\prime }}^{2}b{t}^3+a{t^{\prime }}^2+2bt=0$

Put $t^{\prime }=-t$

$b{t}^3+a{t}^2+2bt=0t(b{t}^2+at+2b)=0b{t}^2+at+2b=0$

For roots to be real $D>0$

${\left(a\right)}^2-4\left(b\right)\left(2b\right)>0a^2-8b^2>0a^2>8b^2$

Hence proved