Question

Find the vertices and foci of the ellipse given by the equation x^2/16+y^2/64=1.

a.) vertices: (8,0),(-8,0); foci: (4/sqrt3, 0),(-4\sqrt3, 0)
b.) vertices: (0,8),(0,-8); foci: (0,4\sqrt3),(0,-4\sqrt3)
c.) vertices: (0,4),(0,-4); foci: (0,4\sqrt3),(0,-4\sqrt3)
d.) vertices: (4,0),(-4,0); foci: (4\sqrt3,0),(-4\sqrt3,0)

Answer 1

Based on the eqn, the eclipse is centered in (0,0) so it should look like this:



so the y-coordinates of all foci n vertices should be 0...which leaves only 2 choices...

Answer 2

which is? it could either be a or d right?

Answer 3

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

\cfrac{x^2}{16}+\cfrac{y^2}{64}=1\implies \cfrac{(x-{\color{red}{ 0}})^2}{4^2}+\cfrac{(y-{\color{red}{ 0}})^2}{{\color{blue}{ 8}}^2}=1\\ \quad \\
\textit{the bigger denominator, or "a", is under the "y", thus is a vertical}\\
\textit{ellipse, thus the foci are from the center, "c" distance over y-coord}\\ \quad \\
center=(h,k)\qquad \qquad foci=({\color{red}{ h,k}}\pm c)\\ \quad \\
c=\sqrt{{\color{blue}{ a}}^2-b^2}\)

Answer 4

Good work :) now we need to find the vertices cuz the foci are the same for a n d...

put in y=0 n solve for x plz

Answer 5

\(\bf vertices = ({\color{red}{ h,k}}\pm {\color{blue}{ a}})\)

Answer 6

I'm so confused.

Answer 7

wel... what's the center of the ellipse?

Answer 8

0?

Answer 9

(0, 25) ? (25, 0) ? (0, -3) ?

Answer 10

I guess I should say what are the (x,y ) coordinates for the center of the ellipse?

Answer 11

I don't know.

Answer 12

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

\cfrac{x^2}{16}+\cfrac{y^2}{64}=1\implies \cfrac{(x-{\color{red}{ 0}})^2}{4^2}+\cfrac{(y-{\color{red}{ 0}})^2}{{\color{blue}{ 8}}^2}=1\\ \quad \\
\textit{the bigger denominator, or "a", is under the "y", thus is a vertical}\\
\textit{ellipse, thus the foci are from the center, "c" distance over y-coord}\\ \quad \\
center=(h,k)\qquad vertices = ({\color{red}{ h,k}}\pm {\color{blue}{ a}})\qquad foci=({\color{red}{ h,k}}\pm c)\\ \quad \\
c=\sqrt{{\color{blue}{ a}}^2-b^2}\)


so... what do you think is the center? the (x,y) coordinates for it that is

Answer 13

(0,-3)
I'm guessing...

Answer 14

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

\cfrac{x^2}{16}+\cfrac{y^2}{64}=1\implies \cfrac{(x-{\color{red}{ 0}})^2}{4^2}+\cfrac{(y-{\color{red}{ 0}})^2}{{\color{blue}{ 8}}^2}=1\qquad center ({\color{red}{ h}}\ ,\ {\color{red}{ k}})\)