Question

If the parabolas $y^2=4b(x-c)$ and $y^2=8ax$ have a common normal except the axis of symmetry of the parabolas, then the range of $\frac{c}{2a-b}$ is

• A. $[0,\infty )$
• B. $(2,\infty )$
• C. $[-2,\infty )$
• D. $[2,\infty )$

Answer 1

The correct option is B $(2,\infty )$

Let $m$ be the slope of the common normal,
Now the equation of normal to parabola $y^2=4b(x-c)$ is
$y=m(x-c)-2bm-b{m}^3\cdots \left(1\right)$

Also, the equation of normal to parabola $y^2=8ax$ is
$y=mx-4am-2a{m}^3\cdots \left(2\right)$

Comparing equations $\left(1\right)$ and $\left(2\right)$, we get
$\frac{-mc-2bm-b{m}^3}{-4am-2a{m}^3}=1$
As the common normal is not the axis of symmetry, so $m\ne 0$,
$\Rightarrow c+2b+b{m}^2=4a+2a{m}^2\Rightarrow c+2b-4a=m^2(2a-b)\Rightarrow \frac{c}{2a-b}-2=m^2$
We know that $m^2>0$, so
$\frac{c}{2a-b}-2>0\Rightarrow \frac{c}{2a-b}>2\therefore \frac{c}{2a-b}\in (2,\infty )$