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 $