I am trying to find the standard form of this equation for an ellipse.
9x^2 + 4y^2 - 36x + 8y + 31 = 0
First I grouped the x and y's together.
9x^2 - 36x + ? + 4y^2 + 8y + ? = -31 + ? + ?
Then I factored the 9 and 4.
9(x^2-4x+4) + 4(x^2+2x+1) = -31+36+4
9(x-2)^2 + 4(x-1)^2 = 9
Then I divided it all by 9 and got:
(x-2)^2 / 1 + 4(x-1)^2 / 9 = 1
Is this correct or did I mess up somewhere?
looks good
other than, you lost the y's, but I think that is just a typo.
you're wrong
You made a slight mistake in your grouping. Let's go through the steps together to find the standard form of the ellipse equation.
Starting with the given equation: 9x^2 + 4y^2 - 36x + 8y + 31 = 0
1. Group the x-terms and y-terms separately:
9x^2 - 36x + 4y^2 + 8y = -31
2. Complete the square for the x-terms:
9(x^2 - 4x) + 4y^2 + 8y = -31
To complete the square, take half of the coefficient of x (-4) and square it: (-4/2)^2 = 4.
Add this value inside the parentheses and compensate by adding it multiplied by 9 outside the parentheses.
Similarly, for the y-terms, take half of the coefficient of y (8) and square it: (8/2)^2 = 16.
Add this value inside the parentheses and compensate by adding it multiplied by 4 outside the parentheses.
9(x^2 - 4x + 4) + 4(y^2 + 2y + 1) = -31 + 9(4) + 4(1)
3. Simplify the equation:
9(x - 2)^2 + 4(y + 1)^2 = -31 + 36 + 4
9(x - 2)^2 + 4(y + 1)^2 = 9
4. Divide the entire equation by the constant on the right side (9) to get the standard form:
(x - 2)^2 / (9/9) + (y + 1)^2 / (9/4) = 1
Simplifying this further:
(x - 2)^2 / 1 + (y + 1)^2 / (9/4) = 1
So, the correct standard form of the given equation is:
(x - 2)^2 + 4(y + 1)^2 / 9 = 1