let equation of an ellipse be x^2+4y^2+6x-8y+9=0
a. Find the standard form of the ellipse
b. Find the center
c. Find the vertices
d. Find the foci
e. Find the eccentricity
x^2+4y^2+6x-8y+9=0
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.
a. To find the standard form of the ellipse, we need to complete the square for both the x and y terms.
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.