Question

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

• A. 44569
• B. 44595
• C. 44563
• D. 44622

Answer 1

The Correct Option is C

Any tangent to $y^{2}= ax$ is $ y= mx+\frac{1}{m} \quad...\left(1\right)$ $ \left[ {\text{Tangent to}} y^{2} = 4x is y = mx +\frac{a}{m}\right] $ $\left(1\right) meets x^{2}= -32y $ where $x^{2}= -32\left(mx+\frac{1}{m} \right)$ $ \Rightarrow x^{2}+32mx +\frac{32}{m} = 0 \quad...\left(2\right)$ Since $\left(1\right)$ touches $x^{2} = -32y$ $\therefore$ roots of $\left(2\right)$ are equal $\therefore \left(32m\right)^{2} = 4\cdot1\cdot\frac{32}{m}$ $ \Rightarrow 8m^{2}= \frac{1}{m}$ $\Rightarrow 8m^{3} $ $ \Rightarrow m^{3} = \frac{1}{8 }$ $\Rightarrow m= \frac{1}{2}$