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
$\Rightarrow mx^{2}-16m^{2}x -8 = 0$ 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} \quad \left[\because m\ne0\right] $
$ \therefore y= -\frac{1}{2} \left(x\right) + \frac{1/2}{-1/2 }$
$ = -\frac{x}{2} -1 $
$ \therefore x+2y+2 = 0$