JEE Main On 8 April 2017 Question 20
Question: 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]
Options:
A) $ 2( \sqrt{2}+1 ) $
B) $ 3+2\sqrt{2} $
C) $ \sqrt{2}+1 $
D) $ 2( 3+2\sqrt{2} ) $
Show Answer
Answer:
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
$ 16m^{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+4m^{2} $
$ m^{4}-4m^{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} $
