Question

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

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

Answer 1

Answer: (A)

$2x+3y+36=0$

A)

$\ln y^2=32xa=8$

and $x^2=108ya=27$

The equation of tangent to the parabola is

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

and also for $x^2=108y$

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

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

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

Now $\Delta =0$

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

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

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

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

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

$\therefore 2x+3y+36=0$