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The line $lx + my + n = 0$is a normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, if
$\frac{{{a^2}}}{{{m^2}}} + \frac{{{b^2}}}{{{l^2}}} = \frac{{({a^2} - {b^2})}}{{{n^2}}}$
$\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$
$\frac{{{a^2}}}{{{l^2}}} - \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} - {b^2})}^2}}}{{{n^2}}}$
None of these
Solution
(b) The equation of any normal to $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is
$ax\sec \theta – $$by {\rm{cosec}}\theta = {a^2} – {b^2}$ …..(i)
The straight line $lx + my + n = 0$…..(ii)
will be a normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
If (i) and (ii) represent the same line
then, $\frac{{a\sec \theta }}{l} = \frac{{b\,{\rm{cosec}}\theta }}{{ – m}} = \frac{{{a^2} – {b^2}}}{{ – n}}$
$ \Rightarrow \cos \theta = \frac{{ – an}}{{l({a^2} – {b^2})}}$ and $\sin \theta = \frac{{bn}}{{m({a^2} – {b^2})}}$
${\cos ^2}\theta + {\sin ^2}\theta = 1$
$\frac{{{a^2}{n^2}}}{{{l^2}{{({a^2} – {b^2})}^2}}} + \frac{{{b^2}{n^2}}}{{m{{({a^2} – {b^2})}^2}}} = 1$
==> $\frac{{{a^2}}}{{{l^2}}} + \frac{{{b^2}}}{{{m^2}}} = \frac{{{{({a^2} – {b^2})}^2}}}{{{n^2}}}$.