Question

The slope of the line touching both the parabolas $y^2=4x$ and $x^2=-32y$ is

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{1}{8}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{1}{2}$

For parabola, $y^2=4x,y=mx+\frac{1}{m}$ is the tangent and it touches the parabola $x^2=-32y$
$\therefore x^2=-32(mx+\frac{1}{m})$
$\Rightarrow x^2+32mx+\frac{32}{m}=0$
Since these parabolas touch only at one point $D=0$
$\Rightarrow (32m{)}^2-4.\left(\frac{32}{m}\right)=0\Rightarrow m^3=\frac{1}{8}\therefore m=\frac{1}{2}$