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 :

Answer 1

Answer:

$m^{2}=2(\sqrt{2}+1)$

Step-by-step explanation:

Hey Mate,

We know that the Standard equation of parabola is :

=  $x^{2}=4ay$

#Figure attach with this answer.

Condition of tangency

=  $c^{2}=a^{2}\; (1+m^{2})$

y=mx+c  is a tangent to the circle $x^{2}+y^{2}=a^{2}$

Tangent $x^{2}+y^{2}=4$ is

= $y=mx\pm 2\sqrt{1+m^{2}}$

And

= $x^{2}=4y$

= $x^{2}=4mx+8\sqrt{1+m^{2}}$

Lets,

D = 0

= $m^{4}-4m^{2}-4=0$

=  $m^{2}=2+2\sqrt{2}$

= $m^{2}=2(\sqrt{2}+1)$