# How do you find the standard form of 9x^2 + 4y^2 - 18x + 16y = 11 and what kind of a conic is it?

Dec 5, 2015

standard form: ${\left(x - 1\right)}^{2} / \left({2}^{2}\right) + {\left(y + 2\right)}^{2} / \left({3}^{2}\right) = 1$
which is an ellipse with center $\left(1 , - 2\right)$, radius along the x-axis of $2$, and radius along the y-axis of $3$

#### Explanation:

Given $9 {x}^{2} + 4 {y}^{2} - 18 x + 16 y = 11$

By regrouping and completing the squares:
$\textcolor{w h i t e}{\text{XXX}} 9 \left({x}^{2} - 2 x + 1\right) - 9 + 4 \left({y}^{2} + 4 y + 4\right) - 16 = 11$

$\textcolor{w h i t e}{\text{XXX}} 9 {\left(x - 1\right)}^{2} + 4 {\left(y + 2\right)}^{2} = 36$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - 1\right)}^{2} / 4 + {\left(y + 2\right)}^{2} / 9 = 1$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - 1\right)}^{2} / \left({2}^{2}\right) + {\left(y + 2\right)}^{2} / \left({3}^{2}\right) = 1$

which is the standard form for an ellipse with the values given in the Answer above.

graph{9x^2+4y^2-18x+16y=11 [-6.22, 6.27, -5.113, 1.127]}