Question

Find the equation of common tangents to the two parabolas $y^2=32x$ and $x^2=108y$

Answer 1

$\begin{array}{c}In{y}^2=3x,a=8\\ andIn{x}^2=108y,a=27\end{array}$

Then equation of tangent to the parabola is

$\begin{array}{c}\Rightarrow y=mx+\frac{a}{m}=mx+\frac{8}{m}\\ andsimilarly\\ \Rightarrow x^2=108y=108\left(mx+\frac{8}{m}\right)\\ \Rightarrow x^2=108mx+\frac{864}{m}\\ \Rightarrow x^{2}m-108m^{2}x-864=0\\ D=0\\ \Rightarrow b^2-4ac=0\\ \Rightarrow {\left(108m^2\right)}^2-4m\left(-864\right)=0\\ \Rightarrow \therefore m=\frac{-2}{3}\end{array}$

$\begin{array}{c}Nowsubsititutingm=\frac{-2}{3}iny=mx+\frac{8}{m}\\ \Rightarrow \therefore y=\frac{-2}{3}x=\frac{8}{\frac{-2}{3}}\\ 2x+3y+36=0.\end{array}$