Question

Given : A circle, $2x^{2}+2y^{2}=5$ and a parabola, $y^{2}=4\sqrt{5}x$
Statement-I : An equation of a common tangent to these curves is $y= x + \sqrt{5}$.
Statement-II : If the line, $y=mx+\frac{\sqrt{5}}{m}\left(m\ne0\right)$ is their common tangent, then $m$ satisfies $m^{4}-3m^{2}+2=0.$

• A. Statement-I is true; statement-II is true; statement-II is a correct explanation for statement-I
• B. Statement-I is true; statement-II is true; statement-II is not a correct explanation for statement-I
• C. Statement-I is true; statement-II is false
• D. Statement-I is false; statement-II, is true

Answer 1

Answer: (B)

Let a tangent to the parabola be $y = mx + \frac{ \sqrt{5}}{m} (m \neq 0)$
As it is a tangent to the circle $x^2 + y^2 = 5/2$, we have $\left(\frac{\sqrt{5}}{m}\right) = \frac{\sqrt{5}}{\sqrt{2}} \sqrt{1+m^{2}} \Rightarrow \left(1+m^{2}\right) m^{2} = 2 $ which gives $m^{4} + m^{2} - 2 = 0 \Rightarrow \left(m^{2} - 1\right)\left(m^{2} + 2 \right) = 0$
As $m \in R , m^2 = 1$ $\therefore \:\: m= \pm 1 $
Also $ m = \pm 1$ does satisfy $m^4 - 3m^2 + 2 = 0$
Hence common tangents are
$ y = x + \sqrt{5} $ and $ y = - x - \sqrt{5}$