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The condition that the straight line $lx + my = n$ may be a normal to the hyperbola ${b^2}{x^2} - {a^2}{y^2} = {a^2}{b^2}$ is given by
$\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{n^2}}}$
$\frac{{{l^2}}}{{{a^2}}} - \frac{{{m^2}}}{{{b^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{n^2}}}$
$\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$
$\frac{{{l^2}}}{{{a^2}}} + \frac{{{m^2}}}{{{b^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$
Solution
(a) Any normal to the hyperbola is
$\frac{{ax}}{{\sec \theta }} + \frac{{by}}{{\tan \theta }} = {a^2} + {b^2}$ …..$(i)$
But it is given by $lx + my – n = 0$…..$(ii)$
Comparing $(i)$ and $(ii),$ we get
$\sec \theta = \frac{a}{l}\left( {\frac{{ – n}}{{{a^2} + {b^2}}}} \right)$
and $\tan \theta = \frac{b}{m}\left( {\frac{{ – n}}{{{a^2} + {b^2}}}} \right)$
Hence eliminating $\theta $, we get
$\frac{{{a^2}}}{{{l^2}}} – \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} + {b^2})}^2}}}{{{n^2}}}$.
