Find the vertices, foci, and eccentricity of the ellipse. Thank you.
x^2/16 + y^2/25 = 1
a^2 = 16 ---> a=4
b^2 = 25 ---> b = 5
major axis along the y-axis,
vertices (4,0) , (-4,0) , (0,5) , (0, -5)
c^ + 4^2 = 5^2
c = ±3
foci : (0,3) , (0, -3)
e = c/b for an ellipse like that
= 3/5
To find the vertices, foci, and eccentricity of the ellipse defined by the equation x^2/16 + y^2/25 = 1, we can first rewrite the equation in standard form.
The standard form for an ellipse centered at the origin is (x^2/a^2) + (y^2/b^2) = 1, where "a" is the horizontal radius and "b" is the vertical radius.
Comparing this with the given equation, we have a^2 = 16 and b^2 = 25.
The vertices of the ellipse are determined by the values of "a" and "b." The vertices lie on the major axis, which is the x-axis in this case.
Therefore, the vertices are (-a, 0) and (a, 0).
Substituting the values, we have (-4, 0) and (4, 0) as the vertices of the ellipse.
The distance from the center to each focus is given by the formula c = sqrt(a^2 - b^2), where "c" is the distance from the center to each focus.
Substituting the values, we have c = sqrt(16 - 25) = sqrt(-9), which is not a real number. This means that the ellipse does not have any real foci.
The eccentricity (e) is calculated as the ratio of the distance from the center to each focus (c) to the value of "a".
Therefore, e = c/a.
Since "c" is not a real number, we can conclude that the eccentricity of this ellipse is not defined.
To summarize:
- The vertices of the ellipse are (-4, 0) and (4, 0).
- The ellipse does not have any real foci.
- The eccentricity of the ellipse is not defined.
To find the vertices, foci, and eccentricity of the ellipse represented by the equation x^2/16 + y^2/25 = 1, we can rewrite the equation in standard form as (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.
Comparing the given equation to the standard form, we can deduce the following values:
Center: (h,k) = (0,0)
a^2 = 16, so a = 4
b^2 = 25, so b = 5
Vertices:
The vertices of an ellipse are given by (±a, k) for a vertically aligned ellipse (when a > b). In this case, the vertices are (±4, 0).
Foci:
The foci of an ellipse can be found using the formula c^2 = a^2 - b^2, where c represents the distance from the center to each focus. Solving for c, we get:
c^2 = 4^2 - 5^2
c^2 = 16 - 25
c^2 = -9
Since we cannot take the square root of a negative number within the real number system, we conclude that this equation represents an imaginary ellipse. Thus, there are no real foci for this particular equation.
Eccentricity:
The eccentricity (ε) of an ellipse can be calculated using the formula ε = c/a, where c is the distance from the center to each focus and a is the semi-major axis. From the previous calculations, we determined that c^2 = -9 and a = 4. Plugging these values into the equation, we get:
ε = (-9/4)
Therefore, the vertices of the ellipse are (±4, 0), there are no real foci, and the eccentricity is -9/4.