A) \[\frac{{{a}^{2}}}{{{l}^{2}}}+\frac{{{b}^{2}}}{{{m}^{2}}}=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{n}^{2}}}\]
B) \[\frac{{{a}^{2}}}{{{m}^{2}}}+\frac{{{b}^{2}}}{{{l}^{2}}}=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{n}^{2}}}\]
C) \[{{a}^{2}}{{l}^{2}}+{{b}^{2}}{{m}^{2}}={{({{a}^{2}}-{{b}^{2}})}^{2}}{{n}^{2}}\]
D) None of the above
Correct Answer: A
Solution :
The equation of any normal to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] is \[ax\,\,\cos \,\theta -by\,\text{cosec }\theta \text{=}{{\text{a}}^{2}}-{{b}^{2}}\] ?. (i) Given equation of straight line is \[lx+my+n=0\] ? (ii) Eqs. (i) and (ii) represent the same line. \[\therefore \] \[\frac{a\,\,\sec \,\theta }{l}=\frac{b\,\text{cosec }\theta }{-m}=\frac{{{a}^{2}}-{{b}^{2}}}{-n}\] Taking Ist a IIIrd terms, IInd and IIIrd terms \[\frac{a\,\,\sec \,\,\theta }{l}=\frac{{{a}^{2}}-{{b}^{2}}}{-n}\] and \[\frac{b\,\text{cosec }\theta }{-m}=\frac{{{a}^{2}}-{{b}^{2}}}{-n}\] \[\Rightarrow \]\[\cos \theta =\frac{-an}{l({{a}^{2}}-{{b}^{2}})}\] and \[\sin \theta =\frac{bn}{m({{a}^{2}}-{{b}^{2}})}\] \[\because \] \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] \[\therefore \] \[\frac{{{a}^{2}}{{n}^{2}}}{{{l}^{2}}{{({{a}^{2}}-{{b}^{2}})}^{2}}}+\frac{{{b}^{2}}{{n}^{2}}}{{{m}^{2}}{{({{a}^{2}}-{{b}^{2}})}^{2}}}=1\] \[\Rightarrow \] \[\frac{{{n}^{2}}}{{{({{a}^{2}}-{{b}^{2}})}^{2}}}\left( \frac{{{a}^{2}}}{{{l}^{2}}}+\frac{{{b}^{2}}}{{{m}^{2}}} \right)=1\] \[\Rightarrow \] \[\frac{{{a}^{2}}}{{{l}^{2}}}+\frac{{{b}^{2}}}{{{m}^{2}}}=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{n}^{2}}}\]
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