What is the common tangent to the parabolas y^2=4ax and x^2=4by ?

1 Answer
Jan 18, 2018

root(3)by+root(3)ax+root(3)(a^2b^2)=0

Explanation:

Let the common tangent be y=mx+c

Let us find the points of intersection of this with parabola y^2=4ax. These are given by

(mx+c)^2=4ax or m^2x^2+x(2mc-4a)+c^2=0

and as it is a tangent, we should have discriminant zero i.e.

(2mc-4a)^2-4m^2c^2=0 or 16a^2-16amc=0 or a=mc (A)

Similarly tangency of y=mx+c with x^2=4by leads to points of intersection given by

x^2-4bmx-4bc=0 and as it is tangent we should have

16b^2m^2+16bc=0 i.e. bm^2+c=0 (B)

Putting value of m from (A) in (B), we get (ba^2)/c^2+c=0 or c^3=-ba^2 i.e. c=-root(3)(ba^2)

and putting this back in (A) we get m=-a/root(3)(ba^2)=-root(3)(a/b)

and hence equation of tangent is y+root(3)(a/b)x+root(3)(ba^2)=0 or root(3)by+root(3)ax+root(3)(a^2b^2)=0

Below we show the graph of parabolas and the tangent, when a=1 and b=2

graph{(y^2-4x)(x^2-8y)(root(3)2y+x+root(3)4)=0 [-5.416, 4.584, -2.8, 2.2]}