Question

Two parabolas $y^2=4a\left(x-l_1\right)and{x}^2=4a\left(y-l_2\right)$ always touch one another then point of contact lies on curve:

• A. $xy=a^2$
• B. $xy=2a^2$
• C. $xy=4a^2$
• D. $none$

Answer 1

Answer: (C)

$xy=4a^2$

$y^2=4a(x-l_1)-----\left(1\right)x^2=4a(y-l_2)-----\left(2\right)$

Let $(\alpha ,\beta )$ be the point of contact.

Explain of tangent to $\left(1\right)$ at $(\alpha ,\beta )$ is

$\beta y=2a(x-l_1+\alpha )\Rightarrow 2ax-\beta y=2a(l_1-\alpha )-----\left(3\right)$

Equation of tangent to $\left(2\right)$ at $(\alpha ,\beta )$ is

$\alpha x=2a(y-l_2+\beta )\Rightarrow \alpha x-2ay=2a(\beta -l_2)-----\left(4\right)$

$\because \left(3\right)$ and $\left(4\right)$ are identical, comparing coefficients of $x$ and $y$ in $\left(3\right)$ and $\left(4\right)$,

$\frac{2a}{\alpha }=\frac{\beta }{2a}\Rightarrow \alpha \beta =4a^2$

$i.e$ the point of contact $(\alpha ,\beta )$ lies on the curve $xy=4a^2$