JEE Main 2017 · 8 Apr · Shift 1 · Mathematics Q80

← JEE Main 2017 · 8 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 80 of 90

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 :

\sqrt2+1

2(3+2\sqrt2)

2(\sqrt2+1)

3+2\sqrt2

Answer and worked solution

Official answer: C

A tangent to $x^2=4y$ is $y=mx-m^2$ . Tangency to the circle needs $\dfrac{m^2}{\sqrt{1+m^2}}=2$ , so $m^4-4m^2-4=0$ and $m^2=2+2\sqrt2$ . The two tangents $y=\pm mx-m^2$ meet at $(0,-m^2)$ , at distance $2(\sqrt2+1)$ from the origin.

Topics:Tangents and normals to conics Tangents and normals

← Previous question Next question →

JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 80