Question

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

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

Answer 1

Answer: (C)

$y^2=4x\dots (i)$
$x^2=-32y\dots (2)$
$m$ be slope of common tangent
Equation of tangent $(1)$
$y=mx+\frac{1}{m}\dots (i)$
Equation of tangent (2)
$y=mx+8m^2\dots (iii)$
(i) and (ii) are identical
$\frac{1}{m}=8m^2$
$\Rightarrow m^3=\frac{1}{8}$
$m=\frac{1}{2}$
Alternative method:
Let tangent to $y^2=4x$ be
$y=mx+\frac{1}{m}$
as this is also tangent to $x^2=-32y$
Solving $x^2+32mx+\frac{32}{m}=0$
Since roots are equal
$\therefore D=0$
$\Rightarrow (32{)}^2-4\times \frac{32}{m}=0$
$\Rightarrow m^3=\frac{4}{32}$
$\Rightarrow m=\frac{1}{2}$