Question

What are the foci of the ellipse given by the equation 100x^2 + 64y^2 = 6 400?

Answer 1

The foci of the ellipse given by the equation 100x^2 + 64y^2 = 6,400 are located at (0, -6) and (0, 6).

Step-by-step explanation:

The foci of the ellipse given by the equation 100x^2 + 64y^2 = 6,400 can be found using the formula c^2 = a^2 - b^2, where c is the distance from the center of the ellipse to each focus, a is the length of the semi-major axis, and b is the length of the semi-minor axis.

In this equation, we can rewrite it in general form, (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h, k) is the center of the ellipse.

By comparing with 100x^2 + 64y^2 = 6,400, we can deduce that the center is (0,0), a^2 = 64, and b^2 = 100. Therefore, a = 8 and b = 10.

Using the formula c^2 = a^2 - b^2, we can calculate c = 6. The foci of the ellipse are located at (0, -6) and (0, 6).

Answer 2

The equation of our ellipse is:

`100 x^2 + 64 y^2 = 6400` (1)

First, let's reduce the equation of the ellipse to the standard form:

`(x^2)/(a^2)+ (y^2)/(b^2)=1`

To do that, we should divide both terms of equation (1) by 6400, and we get:

`(x^2)/(64)+ (y^2)/(100)=1`

This is a vertical ellipse (because
`b^2 \ \textgreater \ a^2`) centered in the origin, and so the distance of its foci from the origin (on the y axis) is given by

`c= √(b^2-a^2)= √((100)-(64))=6`

Therefore, the position of the two foci is (0,6) and (0,-6)