A) \[\frac{{{a}^{2}}}{{{m}^{2}}}+\frac{{{b}^{2}}}{{{l}^{2}}}=\frac{({{a}^{2}}-{{b}^{2}})}{{{n}^{2}}}\]
B) \[\frac{{{a}^{2}}}{{{l}^{2}}}+\frac{{{b}^{2}}}{{{m}^{2}}}=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{n}^{2}}}\]
C) \[\frac{{{a}^{2}}}{{{l}^{2}}}-\frac{{{b}^{2}}}{{{m}^{2}}}=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{n}^{2}}}\]
D) None of these
Correct Answer: B
Solution :
The equation of any normal to \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]is \[ax\sec \theta -\]by \[\text{cosec}\theta ={{a}^{2}}-{{b}^{2}}\] .....(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, \[\frac{a\sec \theta }{l}=\frac{b\,\text{cosec}\theta }{-m}=\frac{{{a}^{2}}-{{b}^{2}}}{-n}\] \[\Rightarrow \cos \theta =\frac{-an}{l({{a}^{2}}-{{b}^{2}})}\] and \[\sin \theta =\frac{bn}{m({{a}^{2}}-{{b}^{2}})}\] \[\because \] \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] \\[\frac{{{a}^{2}}{{n}^{2}}}{{{l}^{2}}{{({{a}^{2}}-{{b}^{2}})}^{2}}}+\frac{{{b}^{2}}{{n}^{2}}}{m{{({{a}^{2}}-{{b}^{2}})}^{2}}}=1\]Þ\[\frac{{{a}^{2}}}{{{l}^{2}}}+\frac{{{b}^{2}}}{{{m}^{2}}}=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{n}^{2}}}\].
You need to login to perform this action.
You will be redirected in
3 sec
