Question

Parabolas $(y-a{)}^2=4a(x-b)$ and $(y-a{)}^2=4a^{\prime }(x-b^{\prime })$ will have a common normal (other than the normal passing through vertex of parabola) if

• A. $2(a-a^{\prime })/(b^{\prime }-b)>1$
• B. $2(a-a^{\prime })/(b-b^{\prime })>1$
• C. $2(a-a^{\prime })/(b+b^{\prime })>1$
• D. $2(a-a^{\prime })/(b^{\prime }+b)>1$

Answer 1

The correct option is A $2(a-a^{\prime })/(b^{\prime }-b)>1$

$(y-a{)}^2=4a(x-b)..................(1)$

$(y-a{)}^2=4a^{\prime }(x-b^{\prime })..............(2)$

Normal to $\left(1\right)$ be,

$y-a=m(x-b)-2am-a{m}^3...............\left(3\right)$

Normal to $\left(2\right)$ be,

$(y-a)=m(x-b^{\prime })-2a^{\prime }m-a^{\prime }{m}^3.................\left(4\right)$

If $\left(1\right)$ and $\left(2\right)$ have common normal, $\left(3\right)$ and $\left(4\right)$ must represent same line, So,

$=>-mb-2am-a{m}^3=-m{b}^{\prime }-2a^{\prime }m-a^{\prime }{m}^3$

$=>(b-b^{\prime })=(2+m^2)(a^{\prime }-a)$

$=>1=(2+m^2)\frac{(a-a^{\prime })}{(b^{\prime }-b)}$

As slope fo Normal must not be zero, so,$(2+m^2)isalwaysgreaterthan$2.$

So, $1<\frac{2(a-a^{p}rime)}{(b^{\prime }-b).}$