Question

Convert the equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci.

Answer 1

Convert the equation to standard form by completing the square on x and y

$100x^2+9y^2+600x+126y+441=0$

collect like terms:

$\begin{gathered} 100x^2+9y^2+600x+126y+441=0 \\ 100x^2+9y^2+600x+126y=-441 \\ 100x^2+600x+9y^2+126y=-441 \end{gathered}$$\begin{gathered} 100x^2+600x+9y^2+126y=-441 \\ 100(x^2+6x)+9(y^2+14y)=-441 \\ 100(x^2+6x+(3)^2)+9(y^2+14y+(7)^2)=-441+100(3)^2+9(7)^2 \\ 100(x+3)^2+9(y+7)^2=-441+900+441 \\ \frac{100(x+3)^2+9(y+7)^2}{900}=\frac{900}{900} \\ \frac{(x+3)^2}{9}+\frac{(y+7)^2}{100}=1 \end{gathered}$$\begin{gathered} a^2=9 \\ a=\sqrt[]{9} \\ a=3 \end{gathered}$$\begin{gathered} b^2=100 \\ b=\sqrt[]{100} \\ b=10 \end{gathered}$$\begin{gathered} c=\sqrt[]{9-100} \\ c=\sqrt[]{-91} \end{gathered}$

The standard form of the ellipse is

$\frac{(x+3)^2}{9}+\frac{(y+7)^2}{100}=1$

While the foci

$\text{foci (-3},-7\pm\sqrt[]{91})$