What are the foci of the ellipse 8x^2+9y^2=96

Step-by-step explanation:

Let's first get this into standard form of an ellipse.  That means that we divide everything by 96 so we have a 1 on the right side of the equals sign:

Doing the division simplifies this down to

We know from the numerators that this is an ellipse centered about the origin (its center is (0, 0).

To find the coordinates of the foci, use

For an ellipse, a is always larger than b, and the larger number is always under the x-term.  So we have a horizontally oriented ellipse.  Our a = 12 and b = 32/3.  Filling in our equation:

and

and

Taking the square root of both sides and simplifying in the process gives you that

So the coordinates of the foci are

47 - 90 = 43

43 - 180 = x

x = 137