Question

The two parabolas $y^2=4a(x-l)$ and $x2=4a(y-l^{\prime })$ always touch one another, the quantities l and l' being both variable; prove that the locus of their point of contact is the curve $xy=4a^2$.

Answer 1

$y^2=4a(x-l).....\left(i\right)x^2=4a(y-l^{\prime }).....\left(ii\right)$

Let the point of contact be $P(h,k)$

Both the parabolas will have a common tangent at their point of contact

Let slope tangent at $P$ to $\left(i\right)$ be $m_1$

$2y\frac{dy}{dx}=4a{m}_1={\left(\frac{dy}{dx}\right)}_{(h,k)}=\frac{2a}{y}=\frac{2a}{k}$

Let slope tangent at $P$ to $\left(ii\right)$ be $m_2$

$2x=4a\frac{dy}{dx}{m}_2{=\left(\frac{dy}{dx}\right)}_{(h,k)}=\frac{x}{2a}=\frac{h}{2a}$

Now $m_1=m_2$

$\frac{2a}{k}=\frac{h}{2a}hk={4a}^2$

Generalising the equation

$xy=4a^2$

Hence proved.