A) \[\frac{{{a}^{2}}}{{{l}^{2}}}-\frac{{{b}^{2}}}{{{m}^{2}}}=\frac{{{({{a}^{2}}+{{b}^{2}})}^{2}}}{{{n}^{2}}}\]
B) \[\frac{{{l}^{2}}}{{{a}^{2}}}-\frac{{{m}^{2}}}{{{b}^{2}}}=\frac{{{({{a}^{2}}+{{b}^{2}})}^{2}}}{{{n}^{2}}}\]
C) \[\frac{{{a}^{2}}}{{{l}^{2}}}+\frac{{{b}^{2}}}{{{m}^{2}}}=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{n}^{2}}}\]
D) \[\frac{{{l}^{2}}}{{{a}^{2}}}+\frac{{{m}^{2}}}{{{b}^{2}}}=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{n}^{2}}}\]
Correct Answer: A
Solution :
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\[\frac{{{a}^{2}}}{{{l}^{2}}}-\frac{{{b}^{2}}}{{{m}^{2}}}=\frac{{{({{a}^{2}}+{{b}^{2}})}^{2}}}{{{n}^{2}}}\].
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