Question

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

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

Answer 1

Answer: (A)

$\frac{a-b}{b}>2$

According to the question...........

$$\begin{array}{c}Parabolasis:\\ y^2=4ax------\left(i\right)\\ eqnofnormal,\\ \Rightarrow y=mx-2am-a{m}^3\\ y^2=4c(x-b)------\left(ii\right)\\ eqnofnormal,\\ \Rightarrow y=m(x-b)-2cm-c{m}^3\\ Now,takevalueofyfromeqn\left(i\right)\&\left(ii\right)\\ y\Rightarrow mx-2am-a{m}^3=m(x-b)-2cm-c{m}^3\\ \Rightarrow mx-2am-a{m}^3=mx-mb-2cm-c{m}^3\\ \Rightarrow 2a+a{m}^2=b+2c+c{m}^2\\ \Rightarrow m^2(a-c)=b+2c-2a\\ \Rightarrow m^2=\frac{b+2c-2a}{a-c}\\ \Rightarrow m^2=\frac{b}{a-c}+\frac{2(c-a)}{(a-c)}=\frac{b}{a-c}-2\\ \Rightarrow m=\sqrt{\frac{b}{a-c}-2}\\ \therefore \frac{b}{a-c}-2>0\\ Commonnormal=\frac{b}{a-c}>2.\\ Soweprovethatbothparabolashavecommonnormalotherthanthex-axis.\end{array}$$

and the correct option is B.