A) \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]
B) \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=2\]
C) \[{{a}^{2}}{{x}^{2}}+{{b}^{2}}{{y}^{2}}=1\]
D) \[{{x}^{2}}-{{y}^{2}}={{a}^{2}}-{{b}^{2}}\]
Correct Answer: B
Solution :
\[\frac{x}{a}\,\cos \theta +\frac{y}{b}sin\theta =1\] ?..(i) \[\frac{x=1}{a}\,\sin \theta -\frac{y}{b}\cos \theta \] ?..(ii) Squaring and adding (i) and (ii), we get \[\Rightarrow \] \[\frac{{{x}^{2}}}{{{a}^{2}}}[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta ]+\frac{{{y}^{2}}}{{{b}^{2}}}[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta ]\] \[+\frac{2x}{a}\frac{y}{b}\cos \theta \sin \theta -\frac{2x}{a}\frac{y}{b}\cos gq\sin \theta =2\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=2\]
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