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, $y=mx+\frac{\sqrt{5}}{m}(m\ne 0)$ 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 not the correct explanation of Statement-I.
• B. Statement-I is true; Statement-II is false.
• C. Statement-I is false; Statement-II is true
• D. Statement-I is true; Statement-II is true; Statement-II is the correct explanation of Statement-I.

Answer 1

The correct option is B Statement-I is true; Statement-II is false.

Let the tangent to the parabola be $y=mx+\frac{\sqrt{5}}{m}(m\ne 0)$ .
Now, its distance from the centre of the circle must be equal to the radius of the circle.
$So,|\frac{\sqrt{5}}{m}|=\frac{\sqrt{5}}{\sqrt{2}}\sqrt{1+m^2}=(1+m^2)m^2=2\Rightarrow m^4+m^2-2=0.$
$\Rightarrow (m^2-1)(m^2+2)=0\Rightarrow m=\pm 1$
So, the common tangents are $y=x+\sqrt{5}$ and $y=-x-\sqrt{5}$.