Equation of the line touching both parabolas $y^2=4 x$ and $x^2=-32 y$ is
Equation of the line touching both parabolas $y^2=4 x$ and $x^2=-32 y$ is
$x+2 y+4=0$
$2 x+y-4=0$
$x-2 y-4=0$
$x-2 y+4=0$
Solution
$y^2=4 x$ ...(i)
Equation of tangent of (i) is, $y=m x+\frac{1}{m}$, ...(ii)
Also, $x^2=-32 y$ ...(iii)
Equation of tangent of (iii) is
$y=m x-(-8 m)^2 \Rightarrow y=m x+8 m^2$ ...(iv)
From (ii) and (iv), $\frac{1}{m}=8 m^2 \Rightarrow m=\frac{1}{2}$ So, from (ii), $y=\frac{x}{2}+2 \Rightarrow x-2 y+4=0$.