Question

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

Answer 1

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

`(x-h)^2/a^2 + (y-k)^2/b^2 = 1" [1]"`

Domain: `h-a <= x <= h+a`
Range: `k-b <= y <=k+b`

Explanation:

Given: `x^2 + 9y^2 = 25" [2]"`

We want equation [2] to look like equation [1] so let's divide both sides of the equation [2] by 25:

`x^2/25 + (9y^2)/25 = 1" [3]"`

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

`x^2/25 + y^2/(25/9) = 1" [4]"`

Write the denominators as squares:

`x^2/5^2 + y^2/(5/3)^2 = 1" [5]"`

Insert 0s for h and k:

`(x-0)^2/5^2 + (y-0)^2/(5/3)^2 = 1" [6]"`

The domain is: `-5 <= x <= 5`
The range is: `-5/3 <= y <= 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]}