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

The Correct Option is C

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