question_answer

If the common tangents to the parabola \[{{x}^{2}}+4y\]and the circle,\[{{x}^{2}}+{{y}^{2}}=4\]intersect at the point P, then the distance of P from the origin, is:        [JEE Online 08-04-2017]

A) \[2\left( \sqrt{2}+1 \right)\]                       

B) \[3+2\sqrt{2}\]

C) \[\sqrt{2}+1\]                   

D) \[2\left( 3+2\sqrt{2} \right)\]

Correct Answer: D

Solution :

 tangent to\[{{x}^{2}}+{{y}^{2}}=4\]                 \[y=mx\pm 2\sqrt{1+{{m}^{2}}}\]                 \[{{x}^{2}}=4y\]                 \[{{x}^{2}}=4mx+8\sqrt{1+{{m}^{2}}}\]                 \[{{x}^{2}}=4mx-8\sqrt{1+{{m}^{2}}}=0\]D = 0                 \[16{{m}^{2}}+4.8\sqrt{1+{{m}^{2}}}=0\]                 \[{{m}^{2}}+2\sqrt{1+{{m}^{2}}}=0\]or\[{{m}^{2}}=\sqrt{1+{{m}^{2}}}\]                 \[{{m}^{4}}=4+4{{m}^{2}}\]                 \[{{m}^{4}}-4{{m}^{2}}-4=0\]                 \[{{m}^{2}}=\frac{4\pm \sqrt{16+16}}{2}\]                 \[=\frac{4\pm 4\sqrt{2}}{2}\]                 \[{{m}^{2}}=2+2\sqrt{2}\]