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 the common tangent, then $m$ satisfies $m^4-3m^2+2=0$.

• A. Statement I is correct, Statement II is correct.
  Statement II is a correct explanation for Statement I
• B. Statement I is correct, Statement II is correct.
  Statement II is not a correct explanation for Statement I
• C. Statement I is correct, Statement II is incorrect
• D. Statement I is incorrect, Statement II is correct.

Answer 1

The correct option is B Statement I is correct, Statement II is correct.

Statement II is not a correct explanation for Statement I
Equation of circle can be rewritten as $x^2+y^2=\frac{5}{2}$
Center $\to$ (0,0) and radius $\to \sqrt{\frac{5}{2}}$
The equation of the tangent to the parabola $y^2=4\sqrt{5}x$ can be written as
$y=mx+\frac{\sqrt{5}}{m}\Rightarrow m^{2}x-my+\sqrt{5}=0$
Since this line is a tangent to the circle as well, the perpendicular from centre to the line is equal to radius of the circle.
$\therefore \frac{\frac{\sqrt{5}}{m}}{\sqrt{1+m^2}}=\sqrt{\frac{5}{2}}\Rightarrow m\sqrt{1+m^2}=\sqrt{2}\Rightarrow m^2(1+m^2)=2\Rightarrow m^4+m^2-2=0\Rightarrow (m^2+2)(m^2-1)=0\Rightarrow m=\pm 1[\because m^2+2\ne 0,asm\in R]$
$\therefore y=\pm (x+\sqrt{5})$, statement I is correct.
$m=\pm 1$ satisfies the given equation of Statement II as $m^4-3m^2+2=0\Rightarrow (m^2-1)(m^2-2)=0$