Question

Find the equations to the common tangents of the parabolas $y^2=4ax$ and $x^2=4by$

Answer 1

The equation of tangent to $y^2=4ax$ is

$y=mx+\frac{a}{m}.......\left(i\right)$

It is also a tangent to $x^2=4by$

$x^2=4b(mx+\frac{a}{m})m{x}^2=4b(m^{2}x+a)m{x}^2-4b{m}^{2}x-4ab=0$

The roots of this equation are equal as it touch the parabola

$\therefore D={(-4b{m}^2)}^2-4m(-4ab)=016b^2{m}^4+16abm=016mb(b{m}^3+a)=0\Rightarrow m=-{\left(\frac{a}{b}\right)}^{\frac{1}{3}}$

Substituting $m$ in $\left(i\right)$, we get

$y=-{\left(\frac{a}{b}\right)}^{\frac{1}{3}}x+\frac{a}{-{\left(\frac{a}{b}\right)}^{\frac{1}{3}}}y=-{\left(\frac{a}{b}\right)}^{\frac{1}{3}}x-a{\left(\frac{b}{a}\right)}^{\frac{1}{3}}y=-{\left(\frac{a}{b}\right)}^{\frac{1}{3}}x-b^{\frac{1}{3}}{a}^{\frac{2}{3}}{b}^{\frac{1}{3}}y=-a^{\frac{1}{3}}x-a^{\frac{2}{3}}{b}^{\frac{2}{3}}{a}^{\frac{1}{3}}x+b^{\frac{1}{3}}y+a^{\frac{2}{3}}{b}^{\frac{2}{3}}=0$