What are the foci of the ellipse given by the equation 225x^2+144y^2=32,400?
To find the foci of the ellipse, we need to first rewrite the equation in standard form:
225x^2 + 144y^2 = 32,400
Divide both sides by 32,400:
225x^2/32,400 + 144y^2/32,400 = 1
Simplify:
x^2/144 + y^2/225 = 1
Now we can identify the center of the ellipse as (0,0), the semi-major axis as sqrt(225) = 15, and the semi-minor axis as sqrt(144) = 12.
The distance from the center to each focus is given by c, which we can find using the formula:
c = sqrt(a^2 - b^2)
where a is the length of the semi-major axis and b is the length of the semi-minor axis. Plugging in the values we get:
c = sqrt(225 - 144) = sqrt(81) = 9
So the foci are located 9 units away from the center along the major axis, at (0, ±9).
Therefore, the foci of the ellipse are (0, 9) and (0, -9).
To find the foci of an ellipse given by its equation, we need to identify the values of a and b, where a is the semi-major axis and b is the semi-minor axis.
First, let's rewrite the equation of the ellipse in standard form:
225x^2 + 144y^2 = 32,400
Divide both sides by 32,400 to simplify:
x^2/144 + y^2/225 = 1
Comparing this to the standard form equation: [x^2/a^2] + [y^2/b^2] = 1, we can see that a^2 = 144 and b^2 = 225.
Taking the square root of both sides, we have a = √144 and b = √225.
Therefore, the values of a and b are a = 12 and b = 15, respectively.
Next, we can find the distance between the foci from the formula: c = √(a^2 - b^2), where c is the distance between the center and the foci.
Using the values of a = 12 and b = 15, we can calculate c:
c = √(12^2 - 15^2)
= √(144 - 225)
= √(-81) (Since 144 - 225 = -81, the result is negative)
The square root of a negative number is imaginary, so there are no real foci for this ellipse.