A) \[{{a}^{2}}{{l}^{2}}+{{b}^{2}}{{m}^{2}}={{n}^{2}}\]
B) \[a{{l}^{2}}+b{{m}^{2}}={{n}^{2}}\]
C) \[{{a}^{2}}l+{{b}^{2}}m=n\]
D) None of these
Correct Answer: A
Solution :
\[y=\frac{-l}{m}x+\frac{n}{m}\]is tangent to \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], if \[\frac{n}{m}=\pm \sqrt{{{b}^{2}}+{{a}^{2}}{{\left( \frac{l}{m} \right)}^{2}}}\] or\[{{n}^{2}}={{m}^{2}}{{b}^{2}}+{{l}^{2}}{{a}^{2}}\].
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