A) \[e=\frac{1}{\sqrt{3}}\]
B) centre is \[(-1,\,2)\]
C) foci are \[(-1,\,1)\]are \[(-1,\,3)\]
D) All of the above
Correct Answer: D
Solution :
We have equation of ellipse is \[3{{x}^{2}}+2{{y}^{2}}+6x-8y+5=0\] \[\Rightarrow \] \[3({{x}^{2}}+2x)+2({{y}^{2}}-4y)+5=0\] \[\Rightarrow \] \[3({{x}^{2}}+2x+1)+2({{y}^{2}}-4y+4)\] \[+5-3-8=0\] \[\Rightarrow \] \[3{{(x+1)}^{2}}+2{{(y-2)}^{2}}=6\] \[\Rightarrow \] \[\frac{{{(x+1)}^{2}}}{2}+\frac{{{(y-2)}^{2}}}{3}=1\] Comparing with \[\frac{{{(x-h)}^{2}}}{{{a}^{2}}}+\frac{{{(y-k)}^{2}}}{{{b}^{2}}}=1,\] we get \[h=-1,\]\[k=2,\]\[{{a}^{2}}=2,\]\[{{b}^{2}}=3\] Now, centre \[\left( h,\,\,k \right)=\left( -1,\text{ }2 \right)\] And using \[{{a}^{2}}={{b}^{2}}(1-{{e}^{2}})\] \[2=3\,(1-{{e}^{2}})\] \[\Rightarrow \] \[e=\frac{1}{\sqrt{3}}\] And foci are \[\left( h,\,\,k+be \right)\] and \[\left( h,\,\,k-be \right)\] \[=\left( -1,\,\,2+1 \right)\]and \[\left( -1,2-1 \right)\] \[=\left( -1,\,\,3 \right)\]and \[\left( -1,\,\,1 \right)\]
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