If the equation

Solution:

Given:

$\left(1+m^{2}\right) x^{2}+2 m c x+\left(c^{2}-a^{2}\right)=0$

Here,

$a=\left(1+m^{2}\right), b=2 m c$ and $c=\left(c^{2}-a^{2}\right)$

It is given that the roots of the equation are equal; therefore, we have:

$D=0$

$\Rightarrow\left(b^{2}-4 a c\right)=0$

$\Rightarrow(2 m c)^{2}-4 \times\left(1+m^{2}\right) \times\left(c^{2}-a^{2}\right)=0$

$\Rightarrow 4 m^{2} c^{2}-4\left(c^{2}-a^{2}+m^{2} c^{2}-m^{2} a^{2}\right)=0$

$\Rightarrow 4 m^{2} c^{2}-4 c^{2}+4 a^{2}-4 m^{2} c^{2}+4 m^{2} a^{2}=0$

$\Rightarrow-4 c^{2}+4 a^{2}+4 m^{2} a^{2}=0$

$\Rightarrow a^{2}+m^{2} a^{2}=c^{2}$

$\Rightarrow a^{2}\left(1+m^{2}\right)=c^{2}$

$\Rightarrow c^{2}=a^{2}\left(1+m^{2}\right)$

Hence proved.