Question

The two parabolas $y^2=4ax$ and $y^2=4c(x-b)$ cannot have a common normal, other than the axis unless, if

• A. $\frac{a-c}{b}>2$
• B. $\frac{b}{a-c}>2$
• C. $\frac{b}{a}+c>2$
• D. $\frac{b}{a-c}<2$

Answer 1

The correct option is D $\frac{b}{a-c}<2$

The two parabolas are given as
$y^2=4ax...\left(1\right)$
and $y^2=4c(x-b)...\left(2\right)$
Equation to any normal to above parabolas will be
$y=mx-2am-a{m}^3...\left(3\right)$
$y=m(x-b)-2cm-c{m}^3...\left(4\right)$
If there is any common normal, then Eqs. $\left(3\right)$ and $\left(4\right)$ must be identical. As the coefficients of $x$ and $y$ are equal, so the constant terms will also be equal, so the constant terms will also be equal, hence
$-2am-a{m}^3=-bm-2mc-c{m}^3$
or $m\left[m^2\right(c-a)-2a+b+2c]=0$.
So, either $m=0$ or $m^2=\frac{2a-b-2c}{c-a}$
If $m=0$, the common normal is the $x-$axis.
If $m^2=\frac{2a-b-2c}{c-a}$,
then $m=\sqrt{\left(\frac{2(a-c)-b}{c-a}\right)}=\sqrt{(-2-\frac{b}{c-a})}$
To have only one common normal, required condition is,
then $-2-\frac{b}{c-a}<0$
$\Rightarrow -\frac{b}{c-a}<2$
$\Rightarrow \frac{b}{a-c}<2$.