Q64 / 200 · Manipal (MET) 2014 · Mathematics
The common tangent of the parabolas $y^2=4 x$ and $x^2=-8 y$ is
- A$y=x+2$
- B$y=x-2$
- C$y=2 x+3$
- Correct answerDNone of these
Solution
The equation of the tangent to parabola $y^2=4 x$ is $y=m x+\frac{1}{m}$
This is also the tangent to parabola $x^2=-8 y$
$\begin{aligned}
& \therefore \quad x^2=-8\left(m x+\frac{1}{m}\right) \\
& \Rightarrow m x^2+8 m^2 x+8=0
\end{aligned}$
It has equal roots, if $\begin{aligned} & 64 m^4 \\ &=32 m \quad\left[\because b^2=4 a c\right] \\ & \Rightarrow \quad m^3=\frac{1}{2} \Rightarrow m=\frac{1}{\sqrt[3]{2}} \end{aligned}$
Hence, equation of the tangent is $y=\frac{1}{\sqrt[3]{2}} x+\sqrt[3]{2}$.
It has equal roots, if $\begin{aligned} & 64 m^4 \\ &=32 m \quad\left[\because b^2=4 a c\right] \\ & \Rightarrow \quad m^3=\frac{1}{2} \Rightarrow m=\frac{1}{\sqrt[3]{2}} \end{aligned}$
Hence, equation of the tangent is $y=\frac{1}{\sqrt[3]{2}} x+\sqrt[3]{2}$.
Hence, the correct answer is option: 'D'.