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 :
Let \[h=\frac{a\,\cos \,t+sin\,t\,+1}{3},\,k=\frac{a\,\sin \,t\,-\,b\,\cos \,t\,+0}{3}\] \[\Rightarrow \,{{(3h-1)}^{2}}+{{(3k)}^{2}}={{a}^{2}}+{{b}^{2}}\] \[\therefore \]Locus of \[\Rightarrow \,2{{x}_{2}}\,+{{y}_{2}}=0\] is \[{{(3x-1)}^{2}}+{{(3y)}^{2}}={{a}^{2}}+{{b}^{2}}\]You need to login to perform this action.
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