JEE Main · 2013 · 07 Apr 2013 · Offline · Parabola

The circle is $x^2+y^2=\frac{5}{2}$ with radius $\sqrt{\frac{5}{2}}$, and for the parabola $y^2=4\sqrt{5}x$ we read off $a=\sqrt{5}$. A slope-form tangent to the parabola is $y=mx+\frac{\sqrt{5}}{m}$. For this line to also touch the circle, the perpendicular distance of the centre from it must equal the radius: $\frac{\sqrt{5}/|m|}{\sqrt{1+m^2}}=\sqrt{\frac{5}{2}}$, which on squaring gives $m^2(1+m^2)=2$, i.e. $m^4+m^2-2=0=(m^2-1)(m^2+2)$. Only $m^2=1$ is admissible, so $m=\pm 1$ and the common tangents are $y=x+\sqrt{5}$ and $y=-x-\sqrt{5}$; hence Statement I is true. Now check Statement II: both admissible slopes $m=\pm 1$ satisfy $m^4-3m^2+2=1-3+2=0$, so every common-tangent slope does satisfy $m^4-3m^2+2=0$ and Statement II is true as well. But that equation factors as $(m^2-1)(m^2-2)=0$ and brings in the extra roots $m=\pm \sqrt{2}$, which are not common tangents, so it is not the tangency condition actually derived. Statement II is therefore true but is not a correct explanation of Statement I.