Question

Prove that two parabolas, having the same focus and their axes in opposite directions, cut at right angles.

Answer 1

Let the equation of first parabola is $y^2=4ax........\left(i\right)$

Then the equtation of the other parabola will be $y^2=-4a(x-2a).........\left(ii\right)$

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

$4ax=-4a(x-2a)\Rightarrow x=a{y}^2=4a\times a\Rightarrow y=\pm 2a$

So the parabolas intersect at $(a,\pm 2a)$

Slope of tangent to $\left(i\right)$ at point of intersection is

$2y\frac{dy}{dx}=4a\frac{dy}{dx}=\frac{2a}{y}{m}_1=\frac{2a}{\pm 2a}=\pm 1$

Slope of tangents to $\left(ii\right)$ at point of intersection

$2y\frac{dy}{dx}=-4a\frac{dy}{dx}=\frac{2a}{-y}{m}_2=\frac{2a}{\pm 2a}=\mp 1$

$m_1{m}_2=(\pm 1)\times (\mp 1)m_1{m}_2=-1$

Hence, they cut each other at right angle.