Question

A normal to the parabola $y^2=4ax$ with slope $m$ touches the rectangular hyperbola $x^2-y^2=a^2$ if :

• A. $m^6+4m^4-3m^2+1=0$
• B. $m^6-4m^4+3m^2-1=0$
• C. $m^6+4m^4+3m^2+1=0$
• D. $m^6-4m^4-3m^2+1=0$

Answer 1

Answer: (C)

$m^6+4m^4+3m^2+1=0$

Given parabola is $y^2=4ax$
Let $m$ be the slope of the normal to the parabola
Equation of normal to the parabola is
$y=mx-2am-a{m}^3$
Since, it touches the hyperbola $x^2-y^2=a^2$
$x^2-(mx-2am-a{m}^3{)}^2=a^2$
$\Rightarrow (1-m^2)x^2+(4a{m}^2+2a{m}^4)x-(a^2+a^2{m}^6+4a^2{m}^2+4a^2{m}^4)=0$
Since, it is tangent to hyperbola
$\Rightarrow (2m^2+m^4{)}^2=(m^2-1\left)\right(1+m^6+4m^2+4m^4)$
$\Rightarrow m^6+4m^4+3m^2+1=0$