Locate the foci of the ellipse. Show your work.
x^2/36+y^2/11=1
c^2=a^2-b^2
c^2=36-11
c^2=25
c= ± sqrt25
c= ±sqrt5*5
is this correct?
correct.
you got c correct, but did not express the coordinates of the foci.
c = ±5
where is the center of the ellipse?
Is the major axis horizontal or vertical?
Yes, your calculation for finding the value of c and the foci of the ellipse is correct. The equation of an ellipse in standard form is given by:
(x^2/a^2) + (y^2/b^2) = 1
where a is the length of the semi-major axis and b is the length of the semi-minor axis.
In your case, the equation is already in standard form:
x^2/36 + y^2/11 = 1
To find the value of c, we use the formula c^2 = a^2 - b^2, where c represents the distance from the center of the ellipse to each focus point.
From the given equation, we can identify that a^2 = 36 and b^2 = 11.
Substituting these values into the formula, we get:
c^2 = 36 - 11
c^2 = 25
Taking the square root of both sides, we find:
c = ± √25
c = ± 5
Therefore, the foci of the ellipse are located at (0, ±5).