A) \[\frac{{{a}^{2}}}{{{(x+h)}^{2}}}-\frac{{{b}^{2}}}{{{(y+k)}^{2}}}=1\]
B) \[\frac{{{a}^{2}}}{{{(x-h)}^{2}}}+\frac{{{b}^{2}}}{{{(y-k)}^{2}}}=1\]
C) \[\frac{{{(x-h)}^{2}}}{{{a}^{2}}}+\frac{{{(y-k)}^{2}}}{{{b}^{2}}}=1\]
D) \[\frac{{{(x-h)}^{2}}}{{{a}^{2}}}-\frac{{{(y-k)}^{2}}}{{{b}^{2}}}=1\]
E) \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]
Correct Answer: B
Solution :
Given equations can be rewritten as \[\cos \theta =\frac{a}{x-n},\sin \theta =\frac{b}{y-k}\] Since, \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] \[\therefore \] \[\frac{{{a}^{2}}}{{{(x-h)}^{2}}}+\frac{{{b}^{2}}}{{{(y-k)}^{2}}}=1\]
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