a. Find the standard form of the ellipse
b. Find the center
c. Find the vertices
d. Find the foci
e. Find the eccentricity
Steve
answered
x^2+6x + 4y^2-8y = -9
x^2+6x+9 + 4y^2-8y+4 = -9+9+4
(x+3)^2 + 4(y-1)^2 = 4
(x+3)^2/4 + (y-1)^1 = 1
Now the rest should be easy. Come on back if you get stuck.
Step-by-Step Bot
answered
Starting with the given equation:
x^2 + 4y^2 + 6x - 8y + 9 = 0
Rearranging terms:
(x^2 + 6x) + 4(y^2 - 2y) + 9 = 0
Completing the square for x:
(x^2 + 6x + 9) + 4(y^2 - 2y) = -9 + 9
(x + 3)^2 + 4(y^2 - 2y) = 0
Completing the square for y:
(x + 3)^2 + 4(y^2 - 2y + 1) = 0 + 4
(x + 3)^2 + 4(y - 1)^2 = 4
Dividing both sides by 4:
(x + 3)^2/4 + (y - 1)^2/1 = 1
The standard form of the ellipse is:
(x + 3)^2/4 + (y - 1)^2/1 = 1
b. To find the center of the ellipse, we look at the standard form. The center is given by the values (h, k) where (h, k) = (-3, 1). Therefore, the center of the ellipse is (-3, 1).
c. To find the vertices, we need to find the length of the major and minor axes. In the standard form (x - h)^2/a^2 + (y - k)^2/b^2 = 1, the length of the major axis is 2a and the length of the minor axis is 2b.
From the standard form:
(x + 3)^2/4 + (y - 1)^2/1 = 1
We can see that a = 2 and b = 1. Therefore, the length of the major axis is 2a = 2(2) = 4. The vertices are located at a distance of 2a from the center along the x-axis. Since the center is (-3, 1), the vertices are (-3, 1 ± 2a) = (-3, 1 ± 4) = (-3, -3) and (-3, 5).
The vertices of the ellipse are (-3, -3) and (-3, 5).
d. To find the foci, we need to find the value of c. In the standard form, c^2 = a^2 - b^2.
From the standard form:
(x + 3)^2/4 + (y - 1)^2/1 = 1
We can see that a = 2 and b = 1. Therefore, c^2 = 2^2 - 1^2 = 4 - 1 = 3. Taking the square root of both sides, c = √3.
The foci of the ellipse are located a distance of c from the center along the major axis. Since the center is (-3, 1), the foci are (-3, 1 ± √3) = (-3, 1 ± 1.732).
The foci of the ellipse are approximately (-3, -0.732) and (-3, 2.732).
e. The eccentricity of an ellipse is given by the formula e = c/a, where c is the distance from the center to the foci and a is the length of the semi-major axis.
From the given values, c = √3 and a = 2.
Therefore, the eccentricity e = (√3)/2 = 0.866 (rounded to three decimal places).
The eccentricity of the ellipse is approximately 0.866.
