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 $\frac{1}{2}$

Let slope of tangent be $m$


Equation of tangent in terms of slope $m$ for parabola $y^2=4x$ is
$y=mx+\frac{1}{m}\cdots \left(1\right)$
This is also a tangent to $x^2=-32y$, so
$\Rightarrow x^2=-32(mx+\frac{1}{m})\Rightarrow m{x}^2+32m^{2}x+32=0$
From the condition of tangency, we get
$D=0\Rightarrow (32m^2{)}^2-4\times m\times 32=0\Rightarrow 32m^4-4m=0\Rightarrow 4m(8m^3-1)=0\Rightarrow m=0\text{or}m=\frac{1}{2}$
$m=0$ (not possible because it is not a tangent to parabola $y^2=4ax$)
$\therefore m=\frac{1}{2}$


Alternate solution :
Equation of tangent in terms of slope $m$ for parabola $y^2=4x$ is
$y=mx+\frac{1}{m}\cdots \left(1\right)$
Now for this line to be tangent to parabola $x^2=-32y$
$\Rightarrow \frac{1}{m}=8m^2$
$\Rightarrow m^3=\frac{1}{8}$
$\Rightarrow m=\frac{1}{2}$