Find the vertices, foci, eccentricity, and length of the latus rectum of the eclipse whose equation is x^2 + 9y^2 = 9
Find the vertices, foci, eccentricity, and length of the latus rectum of the ellipse whose equation is x^2 + 9y^2 = 9
Centre(0,0) divide the equ by 9.frm xsqr/9+ysqr/1=1 asqr:9 a-3 bsqr:1 b-1 vertices:(3,0)(-3,0)(0,1)(0,-1) eccentricity:e-sqrt(1-1/9)=2sqrt2/3 foci=(ae,0)(ae,0)=(2sqrt2,0)(-2sqrt2,0)
To find the vertices, foci, eccentricity, and length of the latus rectum of the ellipse given by the equation x^2 + 9y^2 = 9, we can first rearrange the equation to get it in standard form:
x^2 9y^2 9
Dividing both sides of the equation by 9:
x^2/9 + y^2/1 = 1
Now we can see that the equation is of the form x^2/a^2 + y^2/b^2 = 1, where a^2 = 9 and b^2 = 1. Since a > b, we have an ellipse that is elongated along the x-axis.
The center of the ellipse is given by (h, k), where h is the x-coordinate and k is the y-coordinate. In this case, h = 0 and k = 0. Therefore, the center of the ellipse is (0,0).
The distance from the center to the vertices is given by a, so the vertices are located at (±a, 0). In this case, a = √9 = 3. Therefore, the vertices are (3, 0) and (-3, 0).
The distance from the center to the foci is given by c, where c = √(a^2 - b^2). In this case, c = √(9 - 1) = √8. Therefore, the foci are located at (±√8, 0).
The eccentricity of the ellipse is defined as e = c/a. In this case, e = √8 / 3.
The length of the latus rectum is given by 2b^2/a. In this case, b^2 = 1 and a = 3. Therefore, the length of the latus rectum is 2(1)/3 = 2/3.
So, the vertices are (3, 0) and (-3, 0), the foci are (±√8, 0), the eccentricity is √8 / 3, and the length of the latus rectum is 2/3.