Question

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

• A. $\frac{a-c}{b}>2$
• B. $\frac{b}{a-c}>2$
• C. $\frac{b}{a+c}>2$
• D. None of these

Answer 1

Answer: (B)

Equation to any normal is
$y=mx-2am-a{m}^3$
Equation to any normal is
$y=m(x-b)-2cm-c{m}^3$
If there is any common normal, then they must be identical. As the coefficients of $x$ and y are equal so the constant terms will be also equal, hence
$-2am-a{m}^3=-bm-2mc-c{m}^3$
or $m[m^2(c-a)-2a+b+2c]=0$
So, either $m=0$ or $m^2=\frac{2a-b-2c}{c-a}$
Then, $m=\sqrt{\left(\frac{2(a-c)-b}{c-a}\right)}=\sqrt{\left(-2-\frac{b}{c-a}\right)}$
If the value of m is real and not zero, then
$-2-\frac{b}{c-a}>0,\frac{-b}{c-a}>0$ or $\frac{b}{a-c}>2$