# How do you graph x^2+9y^2=25 and what are the domain and range?

Jan 13, 2017

The domain and range is obvious, when you write the given equation in the standard form for an ellipse:

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1 \text{ }$

Domain: $h - a \le x \le h + a$
Range: $k - b \le y \le k + b$

#### Explanation:

Given: ${x}^{2} + 9 {y}^{2} = 25 \text{ }$

We want equation  to look like equation  so let's divide both sides of the equation  by 25:

${x}^{2} / 25 + \frac{9 {y}^{2}}{25} = 1 \text{ }$

Multiplying the numerator by 9 is the same thing as dividing the denominator by 9:

${x}^{2} / 25 + {y}^{2} / \left(\frac{25}{9}\right) = 1 \text{ }$

Write the denominators as squares:

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

Insert 0s for h and k:

${\left(x - 0\right)}^{2} / {5}^{2} + {\left(y - 0\right)}^{2} / {\left(\frac{5}{3}\right)}^{2} = 1 \text{ }$

The domain is: $- 5 \le x \le 5$
The range is: $- \frac{5}{3} \le y \le \frac{5}{3}$

Here is a graph of the equation

graph{(x-0)^2/5^2 + (y-0)^2/(5/3)^2 = 1 [-10, 10, -5, 5]}