# How do you put x^2-2x+2y^2-12y+3=0 in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

Given:
${x}^{2} - 2 x + 2 {y}^{2} - 12 y + 3 = 0$

Now,
${x}^{2} - 2 x = {x}^{2} - 2 x + 1 - 1$
$= {\left(x - 1\right)}^{2} - 1$
$2 {y}^{2} - 12 y = 2 \left({y}^{2} - 6 y\right)$
$\left({y}^{2} - 6 y\right) = {y}^{2} - 6 y + 9 - 9$
$= {\left(y - 3\right)}^{2} - 9$
$2 {y}^{2} - 12 y = 2 \left({\left(y - 3\right)}^{2} - 9\right)$

The given equation becomes
${\left(x - 1\right)}^{2} - 1 + 2 \left({\left(y - 3\right)}^{2} - 9\right) + 3 = 0$

Simplifying
${\left(x - 1\right)}^{2} - 1 + 2 {\left(y - 3\right)}^{2} - 18 + 3 = 0$
${\left(x - 1\right)}^{2} + 2 {\left(y - 3\right)}^{2} = 18 + 1 - 3$
${\left(x - 1\right)}^{2} + 2 {\left(y - 3\right)}^{2} = 16$

Dividing throughout by 16, we have
${\left(x - 1\right)}^{2} / 16 + 2 {\left(y - 3\right)}^{2} / 16 = \frac{16}{16}$
${\left(x - 1\right)}^{2} / {4}^{2} + {\left(y - 3\right)}^{2} / \left(\frac{16}{2}\right) = 1$
${\left(x - 1\right)}^{2} / {4}^{2} + {\left(y - 3\right)}^{2} / {\left(2 \sqrt{2}\right)}^{2} = 1$

Here,
the major axis is $a = 4$
the minor axis is $b = 2 \sqrt{2}$

Centre is $\left(1 , 3\right)$

End points:
$\left(5 , 3\right)$
$1 , 3 + 2 \sqrt{2}$
$\left(- 3 , 3\right)$
$\left(1 , 3 - 2 \sqrt{2}\right)$

Eccentricity is
$e = \sqrt{\frac{{a}^{2} - {b}^{2}}{a} ^ 2}$
${a}^{2} = 16$
${b}^{2} = 8$
$\sqrt{\frac{{a}^{2} - {b}^{2}}{b} ^ 2} = \sqrt{\frac{16 - 8}{16}}$
$= \sqrt{\frac{8}{16}}$

$e = 0.707$