# How do you write the equation of the ellipse 9x^2 + 4y^2 – 72x + 40y + 208 = 0in standard form?

Nov 15, 2016

The standard form of the ellipse is $\frac{{\left(x - 4\right)}^{2}}{2} ^ 2 + \frac{{\left(y + 5\right)}^{2}}{3} ^ 3 = 1$

#### Explanation:

Let' rewrite the equation step by step

$9 {x}^{2} - 72 x + 4 {y}^{2} + 40 y + 208 = 0$

$9 \left({x}^{2} - 8 x\right) + 4 \left({y}^{2} + 10 y\right) = - 208$

$9 \left({x}^{2} - 8 x + 16\right) + 4 \left({y}^{2} + 10 y + 25\right) = - 208 + 144 + 100$

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

Divide by $36$

$\frac{9 {\left(x - 4\right)}^{2}}{36} + \frac{4 {\left(y + 5\right)}^{2}}{36} = 1$

$\frac{{\left(x - 4\right)}^{2}}{4} + \frac{{\left(y + 5\right)}^{2}}{9} = 1$

graph{9x^2-72x+4y^2+40y+208=0 [-7.69, 14.82, -11.84, -0.59]}