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

1 Answer
Jan 13, 2017

Answer:

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]}