Write the equation in standard form of the ellipse with foci (8, 0) and (-8, 0) if the minor axis has y-intercepts of 2 and -2.
is the answer: x/64+y/4=1
Yes. Except the x and y in the numerators are squared.
okay thanks!
To find the equation of an ellipse in standard form, we first need to determine its center, major axis, and minor axis lengths.
In this case, we are given the foci of the ellipse, which are (8, 0) and (-8, 0). The center of the ellipse is the midpoint between the foci, which is (0, 0).
The minor axis has y-intercepts of 2 and -2. The y-intercepts of an ellipse occur at the endpoints of its minor axis. Thus, the length of the minor axis is 2 units.
The major axis is perpendicular to the minor axis and passes through the center of the ellipse. Since the major axis is not explicitly given in this question, we need to determine its length based on the given information.
The distance between the foci of an ellipse is equal to 2c, where c is the distance from the center to each focus. In this case, c = 8. Thus, the distance between the foci is 2(8) = 16 units.
Since the minor axis has a length of 2 units, the major axis has a length of 16 units + 2 units + 16 units = 34 units.
Now that we have the center (0, 0) and the lengths of the major and minor axes, we can write the equation of the ellipse in standard form:
(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1
where (h, k) is the center of the ellipse, a is the length of the semi-major axis, and b is the length of the semi-minor axis.
Plugging in the values, we have:
(x - 0)^2 / (17)^2 + (y - 0)^2 / (1)^2 = 1
Simplifying, we get:
x^2 / 289 + y^2 = 1
Multiplying both sides by 289 to eliminate the fraction, we have:
x^2 + 289y^2 = 289
So, the equation in standard form of the ellipse with foci (8, 0) and (-8, 0), and minor axis with y-intercepts of 2 and -2 is: x^2 + 289y^2 = 289.