Question

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

Answer 1

`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]}