Find the vertices, foci, and eccentricity of the ellipse.
64x^2 + 81y^2 = 5184
To find the vertices, foci, and eccentricity of the ellipse, we need to rewrite the equation of the ellipse in standard form, which is (x-h)^2/a^2 + (y-k)^2/b^2 = 1 where (h, k) represents the center of the ellipse, and a and b represent the semi-major and semi-minor axes, respectively.
Let's start by rewriting the given equation:
64x^2 + 81y^2 = 5184
To make the coefficients of x^2 and y^2 equal to 1, we divide the entire equation by 5184:
(64x^2)/5184 + (81y^2)/5184 = 1
Now we can rewrite this as:
(x^2/81) + (y^2/64) = 1
Comparing this equation with the standard form, we can see that:
a^2 = 81
b^2 = 64
Taking the square root of both sides:
a = √81 = 9
b = √64 = 8
Therefore, the semi-major axis (2a) is 18, and the semi-minor axis (2b) is 16.
Now, we need to find the center of the ellipse, which is given by the values of (h, k) in the standard form equation. In this case, our center is the origin (0, 0) since there is no h and k in the given equation.
Next, we can find the coordinates of the vertices, which lie on the major axis (the x-axis). The vertices are given by (±a, 0). In this case, it will be:
V1 = (-9, 0)
V2 = (9, 0)
To find the foci of the ellipse, we can utilize the relation c^2 = a^2 - b^2, where c represents the distance from the center to each focus. The foci of the ellipse are given by the coordinates (±c, 0). Substituting the values of a and b:
c^2 = 81 - 64
c^2 = 17
c = √17
Therefore, the coordinates of the foci are:
F1 = (-√17, 0)
F2 = (√17, 0)
Lastly, we can find the eccentricity (e) of the ellipse, which is defined as e = c/a. Substituting the values of c and a:
e = √17/9
The vertices of the ellipse are V1 = (-9, 0) and V2 = (9, 0).
The foci of the ellipse are F1 = (-√17, 0) and F2 = (√17, 0).
The eccentricity of the ellipse is e ≈ 0.564
We have to bring in into form: x^2/a^2+y^2/b^2=1. So dividing everyone by 5184 and simplifying: x^2/81+y^2/64=1
so; x^2/9^2+y^2/8^2=1 therefore a=9 and b=8. Vertices on x-axis (9,0)&(-9,0) on y-axis (0,8)&(0,-8). foci: (c,0)&(-c,0) where c^2=a^2-b^2 so c^2=81-64=17 so c= sqrt71. Locus: (sqrt71,0)&(-sqrt71,0)
eccentricity e=c/a so e= sqrt71/9
try to do the others yourself