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, $y=mx+\frac{\sqrt{5}}{m}\left(m\ne 0\right)$ 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 $y=mx+\frac{\sqrt{5}}{m}(m\ne 0)$
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 $y=-x-\sqrt{5}$