What are the foci of the ellipse given by the equation 225x^2+144y^2=32,400?

Answer 1

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).