what are the x-intercepts, and the y-intercepts and the foci of the ellipse
x^2/25 + y^2/36 = 1
a = 5, b = 6,
the major axis is along the y-axis
so x-intercepts (5,0) (-5,0)
y-intercepts (0,6) (0.-6)
Notice that the denominators are perfect squares. So, either remember sqrts or just substitute in zeroes to get intercepts. Reiny did not give you the foci for some reason.
To find the x-intercepts and y-intercepts of an ellipse, we need to set either x or y to zero and solve for the other variable.
For the given ellipse equation, x^2/25 + y^2/36 = 1, let's find the x-intercepts by setting y = 0:
x^2/25 + 0^2/36 = 1
x^2/25 = 1
x^2 = 25
x = ±√25
x = ±5
So, the x-intercepts are x = ±5.
Now, let's find the y-intercepts by setting x = 0:
0^2/25 + y^2/36 = 1
y^2/36 = 1
y^2 = 36
y = ±√36
y = ±6
Therefore, the y-intercepts are y = ±6.
To find the foci of the ellipse, we need to use the formula c^2 = a^2 - b^2, where a and b are the semi-major and semi-minor axes of the ellipse, and c is the focal length.
In the given equation, a^2 = 25, and b^2 = 36, so:
c^2 = 25 - 36
c^2 = -11
Since c^2 is negative, the ellipse doesn't have any real foci.