What is the major axis and the domain of the equation 9x^2 + 16y^2 - 36x + 96y + 36 = 0 ?
To find the major axis and the domain of the given equation, let's first rewrite it in standard form.
9x^2 + 16y^2 - 36x + 96y + 36 = 0
Rearranging the terms, we have:
9x^2 - 36x + 16y^2 + 96y + 36 = 0
Next, let's complete the square separately for x and y.
For x:
1. Divide every term by the coefficient of x^2, which is 9.
(9x^2 - 36x) / 9 + 16y^2 + 96y + 36 = 0
x^2 - 4x + 16y^2 + 96y + 36 = 0
2. To complete the square, take half of the coefficient of x (-4) and square it: (-4/2)^2 = 4.
Add 4 inside the parentheses, and balance the equation by subtracting 4 outside the parentheses.
x^2 - 4x + 4 + 16y^2 + 96y + 36 - 4 = 0
x^2 - 4x + 4 + 16y^2 + 96y + 32 = 0
3. Simplify the right side by combining like terms.
x^2 - 4x + 16y^2 + 96y + 36 = 0
Now, let's do the same for y:
1. Divide every term by the coefficient of y^2, which is 16.
x^2 - 4x + (16y^2 + 96y + 36) / 16 = 0
x^2 - 4x + y^2 + 6y + 9/4 = 0
2. To complete the square, take half of the coefficient of y (6) and square it: (6/2)^2 = 9.
Add 9 inside the parentheses, and balance the equation by subtracting 9/4 outside the parentheses.
x^2 - 4x + 9/4 + y^2 + 6y + 9/4 - 9/4 = 0
x^2 - 4x + 9/4 + y^2 + 6y + 0 = 0
x^2 - 4x + 9/4 + y^2 + 6y = 0
Now, we have the equation in standard form, which is:
(x - 2)^2 + (y + 3)^2 = 1
Comparing this equation to the standard form of a circle:
(x - h)^2 + (y - k)^2 = r^2
We can see that the center of the circle is at (2, -3) and the radius is 1. Therefore, the major axis of the equation is the diameter of the circle, which passes through the center.
The domain of the equation is the set of all possible x-values, which is (-∞, ∞).
To identify the major axis and domain of the equation 9x^2 + 16y^2 - 36x + 96y + 36 = 0, we need to convert it into a standard form of an ellipse equation, which is:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
The standard form indicates that the center of the ellipse is located at the point (h,k). By completing the square for both the x and y terms, we can rewrite the equation as follows:
9(x^2 - 4x) + 16(y^2 + 6y) = -36
Next, we need to find appropriate values to add and subtract inside the parentheses to complete the square. For the x-term, the appropriate value is (4/2)^2 = 4, and for the y-term, it is (6/2)^2 = 9.
Therefore, we can rewrite the equation as:
9(x^2 - 4x + 4) + 16(y^2 + 6y + 9) = -36 + 36 + 144
Simplifying further:
9(x - 2)^2 + 16(y + 3)^2 = 144
To obtain the standard form, we divide both sides by 144:
(x - 2)^2/16 + (y + 3)^2/9 = 1
Comparing this with the standard form equation, we can conclude that the center of the ellipse is located at the point (2,-3).
The major axis of the ellipse is the longer axis, meaning the axis along which the ellipse stretches the most. In this case, since the denominator under the (x - 2)^2 term is larger than the denominator under the (y + 3)^2 term, the major axis is along the x-axis.
The domain of the ellipse represents all possible x-values that lie on the ellipse. In this case, since the center of the ellipse is at (2,-3) and the major axis is along the x-axis, the domain of the ellipse would extend horizontally from the center to both sides indefinitely.
Hence, the major axis is the x-axis, and the domain is all real numbers.