1 \(e=\frac{1}{\sqrt{3}}\)
2 center is \((-1,2)\)
3 foci are \((-1,1)\) and \((-1,3)\)
4 All of the above
Explanation:
D
The equation of ellipse is
\(3 x^2+2 y^2+6 x-8 y+5=0\)
\(3\left(x^2+2 x\right)+2\left(y^2-4 y\right)+5=0\)
\(3\left(x^2+2 x+1\right)+2\left(y^2-4 y+4\right)+5-3-8=0\)
\(3(x+1)^2+2(y-2)^2=6\)
\(\frac{(x+1)^2}{2}+\frac{(y-2)^2}{3}=1\)
Comparing with
\(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \text {, we get - }\)
\(\mathrm{h}=-1, k=2, a^2=2, b^2=3\)
Here, centre \((\mathrm{h}, \mathrm{k})=(-1,2)\)
And using \(\mathrm{a}^2=\mathrm{b}^2\left(1-\mathrm{e}^2\right)\)
\(2=3\left(1-\mathrm{e}^2\right) \Rightarrow \mathrm{e}=\frac{1}{\sqrt{3}}\)
And foci are ( \(\mathrm{h}, \mathrm{k}+\mathrm{be})\) and ( \(\mathrm{h}, \mathrm{k}-\mathrm{be})\)
\(=(-1,2+1)\) and \((-1,2-1) \quad\binom{\because b e=\sqrt{3} \times \frac{1}{\sqrt{3}}}{b e=1}\)
\(=(-1,3)\) and \((-1,1)\)