The correct option is C 2y2=27x
Given a parabola ,y2=12x
There is a point P outside the parabola with coordinates(h,k).Now the equation of a tangent to the parabola y2=4ax is given by
y=mx+am;m is the slope of tangent
is the equation of any tangent to the parabola y2=12x.This tangent passes through point (h,k) hence
So we can see that there are two possible values of ′m′ and hence we can have two tangents drawn to the parabola y2=12x from the point (h,k).
Also the slope of one is twice that of the other.
So say m1 & m2 are the roots of the quadratic equation(i)
Then we have
Using (vi) we can reduce (v) as 2k29h2=3h
Replace k with y & h with x we have