Question

If a normal of slope $m$ to the parabola $y^2=4ax$ touches the hyperbola $x^2-y^2=a^2$, then

• 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

The correct option is D $m^6+{4m}^4+{3m}^2+1=0$

Equation of normal with slope ${}^{\prime }{m}^{\prime }$ to the parabola $y^2=4ax$ is given by,
$y=mx-2am-a{m}^3$
Also this line touches the hyperbola $x^2-y^2=a^2$
thus using condition of tangency to the hyperbola, $c^2=a^2{m}^2-b^2$
$(-2am-a{m}^3{)}^2=a^2(m^2-1)$
$\Rightarrow 4m^2+m^6+4m^4=m^2-1$
$\Rightarrow m^6+4m^4+3m^2+1=0$
Hence, option 'D' is correct.