# How do you find the standard form of x^2+4y^2+10x+24y+45=0?

Nov 29, 2016

${\left(x + 5\right)}^{2} / {4}^{2} + {\left(y + 3\right)}^{2} / {2}^{2} = 1$

#### Explanation:

Re write it as ${x}^{2} + 10 x + 4 {y}^{2} + 24 y + 45 = 0$

${x}^{2} + 10 x + 25 - 25 + 4 \left({y}^{2} + 6 y + 9 - 9\right) + 45 = 0$

${\left(x + 5\right)}^{2} - 25 + 4 \left({y}^{2} + 6 y + 9\right) - 36 + 45 = 0$

${\left(x + 5\right)}^{2} + 4 {\left(y + 3\right)}^{2} - 16 = 0$

${\left(x + 5\right)}^{2} + 4 {\left(y + 3\right)}^{2} = 16$

${\left(x + 5\right)}^{2} / 16 + {\left(y + 3\right)}^{2} / 4 = 1$

${\left(x + 5\right)}^{2} / {4}^{2} + {\left(y + 3\right)}^{2} / {2}^{2} = 1$

This represents en ellipse centered at (-5,-3) with semi-major axis 4 and semi -minor axis 2