A) \[{{(3x+1)}^{2}}+{{(3y)}^{2}}={{a}^{2}}-{{b}^{2}}\]
B) \[{{(3x-1)}^{2}}+{{(3y)}^{2}}={{a}^{2}}-{{b}^{2}}\]
C) \[{{(3x-1)}^{2}}+{{(3y)}^{2}}={{a}^{2}}+{{b}^{2}}\]
D) \[{{(3x+1)}^{2}}+{{(3y)}^{2}}={{a}^{2}}+{{b}^{2}}\]
Correct Answer: C
Solution :
[c] \[x=\frac{a\cos t+b\sin t+1}{3}\] \[\Rightarrow a\cos t+b\sin t=3x-1\] \[y=\frac{a\sin t-b\cos t}{3}\Rightarrow a\sin t-b\cos t=3y\] Squaring & adding, \[{{(3x-1)}^{2}}+{{(3y)}^{2}}={{a}^{2}}+{{b}^{2}}\]
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