Find the center, vertices, foci, and eccentricity of the ellipse.
9x^2 + 4y^2 - 36x + 8y + 31 = 0
I know the center is (2,-1)
For the vertices I had (3,-1)(1,-1) and the foci (3.8,-1)(.20,-1) and e = 1.80 but I think these are wrong.
I replied to that question yesterday.
Do you agree with my equation?
Since b > a , aren't the foci on the y-axis?
I had
b^2 = a^2 + c^2
9/4 = 1 + c^2
5/4 = c^2
c = ± √5/2
then e = c/b = (√5/2)/(3/2) = √5/3 or appr. .745
if e > 1 ,like you had, you would have a hyperbola
Yesterdays post
http://www.jiskha.com/display.cgi?id=1272908984
so the vertices would be
(2, -1+sqrt5/2) (2, -1-sqrt5/2)?
yes
To find the center, vertices, foci, and eccentricity of the ellipse, we need to rearrange the given equation into standard form for an ellipse. The standard form of an ellipse is:
(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1
where (h, k) represents the center of the ellipse, a represents the semi-major axis length, and b represents the semi-minor axis length.
Let's begin by rearranging the given equation:
9x^2 + 4y^2 - 36x + 8y + 31 = 0
Rearranging, we get:
9x^2 - 36x + 4y^2 + 8y = -31
Completing the square for x terms, we add and subtract (36/2)^2 = 324 to the x terms:
9(x^2 - 4x + 4) + 4y^2 + 8y = -31 + 36
Grouping the x terms and y terms:
9(x - 2)^2 + 4(y^2 + 2y) = 5
Dividing throughout by 9 to get the equation in standard form:
(x - 2)^2/[(5/9)] + (y^2 + 2y)/[(5/4)] = 1
Comparing this equation with the standard form equation, we can identify the following values:
Center: (h, k) = (2, -1)
The center gives us the coordinates of the center of the ellipse.
To find the vertices, we can observe the equation and see that the semi-major axis length is the square root of the denominator of the x term, while the semi-minor axis length is the square root of the denominator of the y term. We have:
a = √[(5/9)] = √(5)/3
b = √[(5/4)] = √5/2
Therefore, the vertices can be found by adding or subtracting the length of the semi-major axis from the x-coordinate of the center:
Vertices: (2 + √(5)/3, -1) and (2 - √(5)/3, -1)
To find the foci, we can use the formula:
c = √(a^2 - b^2)
where c is the distance from the center to each focus. Plugging in the values, we get:
c = √[(5/9) - (5/4)] = √[(20/36) - (45/36)] = √(-25/36)
Since the value inside the square root is negative, it means that the ellipse is not defined in the real number plane, and therefore has no real foci.
Finally, the eccentricity of the ellipse can be found using the equation:
e = c / a
Plugging in the values, we get:
e = (√(-25/36)) / (√(5)/3)
Since the value inside the square root is negative, it means that the eccentricity is imaginary, indicating that the ellipse is not defined in the real number plane.
Therefore, the given values for the vertices and foci you provided are incorrect.