A conic section has the equation x^2+y^2+12x+8y=48. Determine the following; type of conic, domain and range, axes of symmetry, and center.
To determine the type of conic, domain, range, axes of symmetry, and center of the given equation x^2+y^2+12x+8y=48, we will need to complete the square for both the x and y variables.
Starting with the x terms:
x^2 + 12x = -y^2 - 8y + 48
To complete the square for the x terms, we take half of the coefficient of x (which is 12), square it, and add it to both sides of the equation:
x^2 + 12x + 36 = -y^2 - 8y + 48 + 36
(x + 6)^2 = -y^2 - 8y + 84
Similarly, for the y terms:
y^2 + 8y = -x^2 - 12x + 48
Completing the square for the y terms:
y^2 + 8y + 16 = -x^2 - 12x + 48 + 16
(y + 4)^2 = -x^2 - 12x + 64
Now let's simplify and rewrite the equation:
(x + 6)^2 + (y + 4)^2 = -x^2 - 12x + 64 + -y^2 - 8y + 84 + 16
x^2 + 12x + 36 + y^2 + 8y + 16 = -x^2 - 12x - y^2 - 8y + 64 + 84 + 16
2x^2 + 2y^2 + 20x + 16y = 180
Now, let's rearrange the terms to better understand the equation:
2x^2 + 20x + 2y^2 + 16y = 180
To determine the type of conic, we can compare the coefficients of x^2 and y^2. Since they have the same coefficient (2) and are positive, this equation represents an ellipse.
The domain and range of the ellipse can be determined based on the range of possible x and y values that satisfy the equation. Since there are no restrictions mentioned in the equation, the domain and range are all real numbers (-∞, ∞).
The axes of symmetry for an ellipse are the lines passing through its center. To find the center, we can rewrite the equation in standard form:
(x + 6)^2 + (y + 4)^2 = 180
Comparing this to the standard form (x-h)^2/a^2 + (y-k)^2/b^2 = 1, we can determine that the center of the ellipse is at the point (-6,-4).
In summary:
- The given equation x^2+y^2+12x+8y=48 represents an ellipse.
- The domain and range are all real numbers (-∞, ∞).
- The axes of symmetry are lines passing through the center.
- The center of the ellipse is (-6, -4).