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: B
Solution :
\[3h=a\cos t+b\sin t+1,\,\,3k=a\sin t-b\cos t\] \[{{a}^{2}}+{{b}^{2}}={{(3h-1)}^{2}}+{{(3k)}^{2}}\] \[{{(3x-1)}^{2}}+{{(3y)}^{2}}={{a}^{2}}+{{b}^{2}}\].You need to login to perform this action.
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