Question

The equation of the common tangent to the parabolas $y^2=2x$ and $x^2=16y$ will be

• A. $x+y+2=0$
• B. $x-3y+1=0$
• C. $x+2y-2=0$
• D. $x+2y+2=0.$

Answer 1

Answer: (D)

The equation of any tangent to $y^2=2x$ is
$y=mx+\frac{1/2}{m}(m\ne 0)(\because a=\frac{1}{2})$
It this line touches $x^2=16y$, then
$x^2=16\left(mx+\frac{1}{2m}\right)$ has equal roots
$\Rightarrow x^2=\frac{8\left(2m^{2}x+1\right)}{m}$ has equal roots
$\Rightarrow m{x}^2-16m^{2}x-8=0$ has equal roots
$\Rightarrow {\left(16m^2\right)}^2-4\left(m\right)\left(-8\right)=0$
$\Rightarrow 256m^4+32m=0$
$\therefore 8m^3+1=0$
$\Rightarrow m=-\frac{1}{2}\left[\because m\ne 0\right]$
$\therefore y=-\frac{1}{2}\left(x\right)+\frac{1/2}{-1/2}$
$=-\frac{x}{2}-1$
$\therefore x+2y+2=0$