Question

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

Answer 1

Answer:

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:

$\frac{8x^2}{96}+\frac{9y^2}{96}=1$

Doing the division simplifies this down to

$\frac{x^2}{12}+\frac{y^2}{\frac{32}{3} }=1$

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

$c^2=a^2-b^2$

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:

$c^2=12^2-(\frac{32}{3})^2$

and

$c^2=144-\frac{1024}{9}$ and

$c^2=\frac{272}{9}$

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

$c=+/-\frac{4\sqrt{17} }{3}$

So the coordinates of the foci are

$(-\frac{4\sqrt{17} }{3},0)(\frac{4\sqrt{17} }{3},0)$