Question

The y-intercept of the common tangent to the parabola $y^2=32x$ and $x^2=108y$ is ?

• A. $-18$
• B. $-12$
• C. $-9$
• D. $-6$

Answer 1

The correct option is B $-12$

Here, equation of the tangent to the parabola $y^2=32x$ is

$y=mx+\frac{8}{x}⟶\left(1\right)[since,4a=32\Rightarrow a=8]$

and it meets $x^2=108y$

So,

$x^2=108(mx+\frac{8}{x})\Rightarrow m{x}^2=108m^{2}x+864\Rightarrow m{x}^2-108m^{2}x-864=0$

If the line $\left(1\right)$ is the tangent to the second parabola, then the roots of the above equation must be equal.

So,

${(-108m^2)}^2-4\left(m\right)(-864)=0[since,b^2-4ac=0]\Rightarrow 27m^3+8=0\Rightarrow m^3=\frac{-8}{27}\Rightarrow m=\frac{-2}{3}$

Substituting the value of $m$ in equation $\left(1\right)$ we get,

$y=-\frac{2x}{3}+\frac{8}{\frac{-2}{3}}\Rightarrow y=\frac{-2x-36}{3}\Rightarrow 3y=-2x-36\Rightarrow 2x+3y+36=0$

This is the equation of the common tangent.

Now, to find the $y$ - intercept, let $x=0$

$\therefore 3y+36=0\Rightarrow 3y=-36\Rightarrow y=-12$

Hence, the required $y$ - intercept is $-12.$