Locate the foci of the ellipse. Show your work.
𝑥2/36+ y2/11=1
I will assume your ellipse is
x^2/36 + y^2/11 = 1
so a = 6 and b = √11 with the foci along the x-axis
for that kind of ellipse,
c^2 = a^2 - b^2
= 36 - 11 = 25
c = ± 5
thus the foci are (5,0) and (-5,0)
Thank you guys :)
If you mean
x^2/36 + y^2/11 = 1
then you can see that since 36 > 11, the major axis is on the x-axis.
So, since b^2+c^2 = a^2, c^2=25, and the foci are at (±5,0)
To locate the foci of an ellipse, we need to find the values of a and b in the equation of the ellipse. The general equation of an ellipse is given by:
(x^2/a^2) + (y^2/b^2) = 1
Comparing this with the given equation, we have:
(x^2/36) + (y^2/11) = 1
Here, a^2 = 36, so a = √36 = 6, and b^2 = 11, so b = √11.
The distance from the center of the ellipse to each focus is c, where c is calculated using the formula:
c^2 = a^2 - b^2
Substituting the values, we can find:
c^2 = 6^2 - √11^2
c^2 = 36 - 11
c^2 = 25
Taking the square root of both sides, we get:
c = √25
c = 5
So the distance from the center of the ellipse to each focus is 5 units.
To locate the foci, we need to consider the center of the ellipse, which in this case is at the origin (0,0). We can find the coordinates of the foci by adding or subtracting c from the x-coordinate of the center.
Therefore, the coordinates of the foci are (6, 0) and (-6, 0).