$[{x^2} + (y - a)(y + a)] + \lambda x = 0 \Rightarrow {x^2} + {y^2} + \lambda x - {a^2} = 0$
and $\sqrt {{{\left( {\frac{\lambda }{2}} \right)}^2} + {a^2}} = \frac{{\frac{{ - m\lambda }}{2} + c}}{{\sqrt {1 + {m^2}} }}$
$ \Rightarrow (1 + {m^2}){\rm{ }}\left[ {\frac{{{\lambda ^2}}}{4} + {a^2}} \right] $
$= {\left( {\frac{{m\lambda }}{2} - c} \right)^2}$
$ \Rightarrow (1 + {m^2}){\rm{ }}\left[ {\frac{{{\lambda ^2}}}{4} + {a^2}} \right] $
$= \frac{{{m^2}{\lambda ^2}}}{4} - mc\lambda + {c^2}$
$ \Rightarrow {\lambda ^2} + 4mc\lambda + 4{a^2}(1 + {m^2}) - 4{c^2} = 0$
$\therefore $${\lambda _1}{\lambda _2} = 4[{a^2}(1 + {m^2}) - {c^2}]$
$\Rightarrow {g_1}{g_2} = [{a^2}(1 + {m^2}) - {c^2}]$
and ${g_1}{g_2} + {f_1}{f_2} = \frac{{{c_1} + {c_2}}}{2}$
$\Rightarrow {a^2}(1 + {m^2}) - {c^2} = - {a^2}$
Hence ${c^2} = {a^2}(2 + {m^2})$.
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