# How do you graph 4x^2+9y^2+24-18+9=0?

Sep 8, 2016

(see below)

#### Explanation:

Note: I am assuming that the equation should have been:
$\textcolor{w h i t e}{\text{XXX}} 4 {x}^{2} + 9 {y}^{2} + 24 \textcolor{red}{x} - 18 \textcolor{red}{y} + 9 = 0$

The standard form of an ellipse is
$\textcolor{w h i t e}{\text{XXX}} \frac{{\left(x - h\right)}^{2}}{{a}^{2}} + \frac{{\left(y - k\right)}^{2}}{{b}^{2}} = 1$

with center at $\left(h , k\right)$,
radius on X-axis $= a$, and
radius on Y-axis$= b$

Converting the (assumed) equation into this form:
$\textcolor{w h i t e}{\text{XXX}} 4 \left({x}^{2} + 6 x\right) + 9 \left({y}^{2} - 2 y\right) = - 9$

$\textcolor{w h i t e}{\text{XXX}} 4 \left({x}^{2} + 6 x + 9\right) + 9 \left({y}^{2} - 2 y + 1\right) = - 9 + 4 \times 9 + 9 \times 1$

$\textcolor{w h i t e}{\text{XXX}} 4 {\left(x + 3\right)}^{2} + 9 {\left(y - 1\right)}^{2} = 36$

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

So we need to draw an ellipse with center at $\left(- 3 , 1\right)$,
a radius of $3$ parallel to the X-axis, and
a radius of $2$ parallel to the Y-axis. 