Question

Find the center, vertices, and foci of the ellipse with equation x^2/400+ y^2 /256= 1

Answer 1

@dan815 can you please help me?

Answer 2

I assume you've covered ellipses already?

Answer 3

yes

Answer 4

\(\bf \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 \\
\cfrac{x^2}{400}+\cfrac{y^2}{256}=1\implies \cfrac{(x-0)^2}{20^2}+\cfrac{(y-0)^2}{16^2}=1\)

see the center?

notice, the bigger denominator is under the "x"
that means the major axis is over the x-axis
so is a horizontal ellipse

so from the center, you move "a" units to the left and right
and you'd find the vertices there

Answer 5

Ok so is the center (-20,0) and (20,0)?

Answer 6

yeap

Answer 7

so, to find the focus points, or foci

well, again, we move from the center, over the x-axis
but this time
a difference distance
we'll move to the left and right by "c" units
and \(\bf c=\sqrt{a^2-b^2}\)

Answer 8

c= sqrt 400-256 =144
144=12^2

Answer 9

well... kinda... just 12 \(\bf c=\sqrt{400-256}\implies c=\sqrt{144}\implies c=12\)

Answer 10

OK great I understand! And how are you supposed to know if the major axis is for x or y if the problem doesn't say?

Answer 11

for ellipses
is defined by the "bigger denominator"

400 is bigger than 256
and "x" is above the 400
so the major axis runs over the x-axis

for hyperbolas
is defined by the "positive fraction"
which is usually at the right-side anyway

so if the positive fraction is the one with the "y" in it
that means the hyperbola transversal axis runs over the y-axis
if it's the "x" fraction
then it runs over the x-axis

Answer 12

hmm well. the positive fraction in hyperbolas, is usually at the left-side rather
usually, doesn't have to, but usually

Answer 13

Ok well thank you for all of your help! :)

Answer 14

yw