Question

The equation of the common tangent of the parabolas $x^2=108y$ and $y^2=32x$, is

• A. $2x+3y=36$
• B. $2x+3y+36=0$
• C. $3x+2y+36=0$
• D. $3x+2y=36$

Answer 1

The correct option is B $2x+3y+36=0$

Given $x^2=108y=4\times 27y\Rightarrow a=27$

and $y^2=32x=4\times 8x\Rightarrow a=8$

The equation of tangent to the parabola is

$y=mx+\frac{a}{m}=mx+\frac{8}{m}$

Also, for $x^2=108y$

$\therefore x^2=108y=108(mx+\frac{8}{m})$

$x^2=108mx+\frac{864}{m}$

$\therefore m{x}^2-108m^{2}x-864=0$

Now $\Delta =0$

$\therefore b^2-4ac=0$

$\therefore {\left(108m^2\right)}^2-4\times m\times -864=0$

$\therefore 27m^3+8=0$

$\therefore m=\frac{-2}{3}$

Now substituting $m=\frac{-2}{3}$ in $mx+\frac{a}{m}$

$\therefore y=\frac{-2}{3}x+\frac{8}{\left(\frac{-2}{3}\right)}$

$\therefore 2x+3y+36=0$ is the required equation.