If the parabolas `y^2=4a x` and `y^2=4c(x-b)` have a common normal other than the x-axis `(a , b , c` being distinct positive real numbers), then prove that `b/(a-c)> 2.`

Options

A. `0ltb/(a-c)lt1`
B. `b/(a-c)gt2`
C. `b/(a-c)lt0`
D. `1ltb/(a-c)lt2`

Answer

B

Explanation

The equation of any normal of slope m to the parabola `y^(2)=4c(x-c)` is
`y=m(x-b)-2cm-cm^(2)or, y=mx-mb-2cm-cm^(3)`
For this to be normal to `y^(2)=4ax`, we must have
`-2am-am^(3)=-mb-2cm-cm^(3)`
`rArr" "2a+am^(2)=b+2c+cm^(2)`
`rArr" "m^(2)=(b+2c-2a)/(a-c)rArrm^(2)=b/(ac)-2rArrm=+-sqrt(b/(a-c)-2)`
For m to be real, we must have `-b/(a-c)-2gt0rArr=b/(a-c)gt2`