Question

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

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

Answer 1

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

Let $y^2=32x$ has a tangent $y=mx+\frac{a}{m}$,........(1)

The equation of the tangent to the parabola $y^2=32x$

$y=mx+\frac{8}{m}$........(2)

If the line given by (2) is also a tangent to the parabola $x^2=108y$,

then eqn. (2) meets $x^2=108y$ ......(3)in two coincident points,

$\Rightarrow$ substituting the value of y from (2) in (3), we get

$\Rightarrow$ $x^2=108[m\cdot x+\frac{8}{m}]$

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

The root of this quadratic equation are equal provided$B^2=4AC$

$\Rightarrow (-108m^2{)}^2=4m(-864)$

$\Rightarrow$ $m=\frac{-2}{3}$ (m ia not equal to zero)

substituting their value of m in (2), the required equation is

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

$y=\frac{-2}{3}\cdot x-12$

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

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