Question
Write the centre and eccentricity of the ellipse $3\text{x}^2+4\text{y}^2-6\text{x}+8\text{y}-5=0.$

Answer

$\Rightarrow3\text{x}^2-6\text{x}+4\text{y}^2+8\text{y}-5=0$
$\Rightarrow3(\text{x}^2-2\text{x})+4(\text{y}^2+2\text{y})=5$
$\Rightarrow3(\text{x}^2-2\text{x}+1)+4(\text{y}^2+2\text{y}+1)=5+3+4$
$\Rightarrow3(\text{x}-1)^2+4(\text{y}+1)^2=12$
$\Rightarrow\frac{3(\text{x}-1)^2}{12}+\frac{4(\text{y}+1)^2}{12}=1$
$\Rightarrow\frac{(\text{x}-1)^2}{4}+\frac{(\text{y}+1)^2}{3}=1$
Compairing it with $\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ we get:
a = 2 and $\text{b}=\sqrt{3}$
Here, a > b, so the major and the minor axes of the ellipse are along the x−axis and y−axis, respectively.
Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{3}{4}}$
$\Rightarrow\text{e}=\sqrt{\frac{1}{4}}$
$\therefore\ \text{e}=\frac{1}{2}$ and centre = (1, -1)

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