Find the vertices, foci, eccentricity and length of the latus rectum of the ellipse whose equation is x^2 + 9Y^2 = 9
divide by 9
x^2/(1/9) + y^2 = 1
a^2 = 1/9, a = ±1/3
b = ±1
so we have an ellipse with the major axis as the y-axis
vertices:
on x-axis ( 1/3,0) , (-1/3 , 0)
on y-axis (0,1) , (0,-1)
foci:
c^2 + (1/3)^2 = 1^1
c^2 = 1-1/9 = 8/9
c= ± (2√2)/3
this will give you the foci
e =c/b= (2√2/3)/1 = (2/3)√2
Use your definition of latus rectum to finish the question.
To find the vertices, foci, eccentricity, and length of the latus rectum of the ellipse with the equation x^2 + 9y^2 = 9, we first need to determine the standard form of the equation.
The standard form of an ellipse equation is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h, k) represents the center of the ellipse, a is the semi-major axis, and b is the semi-minor axis.
Given equation: x^2 + 9y^2 = 9
Dividing both sides by 9, we get: (x^2)/9 + (y^2)/1 = 1
Comparing this with the standard form, we have: (x-0)^2/3^2 + (y-0)^2/1^2 = 1
Therefore, the center of the ellipse is (h, k) = (0, 0), and the semi-major axis, a = 3, and semi-minor axis, b = 1.
Now, let's find the vertices:
The vertices of the ellipse are given by (h, k ± a).
Vertices: (0, 0 ± 3) = (0, ± 3)
So the vertices are (0, 3) and (0, -3).
To find the foci of the ellipse, we can use the formula c = √(a^2 - b^2). The foci points will be (h, k ± c).
c = √(3^2 - 1^2) = √8 = 2√2
Foci: (0, 0 ± 2√2) = (0, ± 2√2)
So the foci are (0, 2√2) and (0, -2√2).
The eccentricity of an ellipse can be calculated using the formula e = c/a.
eccentricity e = (2√2)/3
The length of the latus rectum of an ellipse can be calculated using the formula 2b^2/a.
Length of latus rectum = 2(1^2)/3 = 2/3
Summary:
- Vertices: (0, 3) and (0, -3)
- Foci: (0, 2√2) and (0, -2√2)
- Eccentricity: (2√2)/3
- Length of latus rectum: 2/3
To find the vertices, foci, eccentricity, and length of the latus rectum of the ellipse with the equation x^2 + 9y^2 = 9, we can analyze the general form of an ellipse equation:
(x - h)^2/a^2 + (y - k)^2/b^2 = 1,
where:
- (h, k) represents the coordinates of the center of the ellipse.
- The vertices are at (h ± a, k).
- The foci are at (h ± c, k), where c = sqrt(a^2 - b^2).
- The eccentricity is given by ε = c/a.
- The length of the latus rectum is 2b^2/a.
Comparing this to the given equation x^2 + 9y^2 = 9, we can see that the equation is already in standard form, and we have:
a = 1 (since a^2 = 1),
b = 1/3 (since b^2 = 1/9),
c = sqrt(a^2 - b^2) = sqrt(1 - 1/9) = sqrt(8/9) = (2/3)sqrt(2),
ε = c/a = (2/3)sqrt(2) / 1 = (2/3)sqrt(2),
and the length of the latus rectum = 2b^2/a = 2(1/3)^2 / 1 = 2/9.
Therefore, for the given equation x^2 + 9y^2 = 9:
- The vertices are at (±1, 0).
- The foci are at (±(2/3)sqrt(2), 0).
- The eccentricity is ε = (2/3)sqrt(2).
- The length of the latus rectum is 2/9.