Question

you will get a medal :)
Locate the foci of the ellipse. Show your work.

x^2/36+y^2/11=1

Answer 1

any ideas where the ellipse center is at?

Answer 2

notice your center \(\cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ a}}^2}+\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ b}}^2}=1

\qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad
vertices\ ({\color{brown}{ h}}\pm a, {\color{blue}{ k}})
\\ \quad \\

% ellipse, vertical major axis

\cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ b}}^2}+\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ a}}^2}=1

\qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad
vertices\ ({\color{brown}{ h}}, {\color{blue}{ k}}\pm a)\)

Answer 3

anyhow, get the center, see if it's horizontal or a vertical major axis
then get the "c" distance, "c" is the distance from the center to either foci
and \(\bf c=\sqrt{a^2-b^2}\)

Answer 4

Continuing from above,

c = sqrt(36 - 11)

c = sqrt(25)

c = 5

@nthenic_oftime

Can you get the foci coordinates now.