Question

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

Answer 1

hmmm do you know an ellipse's equation?

Answer 2

could you remind me?

Answer 3

http://www.mathwarehouse.com/ellipse/images/translations/formula_for_equations.gif

Answer 4

\(\bf 100x^2+64y^2=6,400\implies \cfrac{100x^2}{6400}+\cfrac{64y^2}{6400}=1\implies \cfrac{x^2}{64}+\cfrac{y^2}{100}=1\\ \quad \\
\cfrac{x^2}{8^2}+\cfrac{y^2}{10^2}=1\implies \cfrac{(x-0)^2}{8^2}+\cfrac{(y-0)^2}{10^2}=1\)

Answer 5

the bigger denominator is under the "y", so that means the ellipse is moving vertically

so the foci will be \(\bf "c"\) distance from the center of it, over the y-axis

\(\bf c=\sqrt{a^2-b^2}\)

Answer 6

"a" component is always the BIGGER denominator in an ellipse
"b" is the smaller one

Answer 7

anyhow, to avoid any ambiguities from the squared form
a = 10 b = 8

Answer 8

\[0,\pm36\]

Answer 9

is that correct
i followed you formula

Answer 10

close, \(\bf c = \sqrt{a^2-b^2}\implies c=\sqrt{100-64}\implies c = \sqrt{36}\)

Answer 11

so the \[\pm36\] comes first?

Answer 12

nope, you're correct, you just forgot to take square root of 36

Answer 13

\(\bf c = \sqrt{a^2-b^2}\implies c=\sqrt{100-64}\implies c = \sqrt{36}\\ \quad \\
c = 6\qquad \qquad (0,\pm 6)\)

Answer 14

ohhhhhh gotchaaaaaa