Question

find a set of parametric equations for the line or conic.
Ellipse: Vertices: (+/-5,0) & Foci: (+/-4,0)

Answer 1

do you know what the center is?

Answer 2

no it provides us with that

Answer 3

the center of the ellipse is the two foci isn't it?

Answer 4

right i know,
my question really meant "given the information, what is the center?"

Answer 5

no there is one center for an ellipse
two foci, but one center
it is half way between the foci, so in this case it is right at the origin \((0,0)\)

Answer 6

the general form of an ellipse is
\[\frac{(x-h)^2}{a^2}+\frac{(y-h)^2}{b^2}=1\] but in your example, since the center is \((0,0)\) it is
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] and your job now is to find \(a\) and \(b\)

Answer 7

there asking for two parametric equations and would would put them into these
x-h+acos theta
y=k+bcos theta
i just don't know what to plug in where

Answer 8

oh i see
you can ignore the \(h\)and \(k\) part because the center is \((0,0)\) and so \(h=k=0\)

Answer 9

so it is just going to be
\[x=a\cos(\theta)\] and \[y=b\sin(\theta)\]

Answer 10

since the ellipse travels through the point \((5,0)\) you know it is
\[x=5\cos(\theta)\]

Answer 11

does the negative have any influence?

Answer 12

that way when \(\theta=0\) you get \(5\cos(0)=5\) and \(5\sin(0)=0\)

Answer 13

no not really \(\theta\) will go form \(0\) to \(2\pi\) so you will get negative values that way
the only work you need to do here is to find where the ellipse hits the \(y\) axis so you can put that number for \(b\) in \(b\sin(\theta)\)

Answer 14

and you do that by noting that \(a^2=b^2+c^2\) or in your case
\[5^2=b^2+4^2\] making
\[b=3\]

Answer 15

your parametric equations are therefore
\[x=5\cos(\theta), y=3\sin(\theta)\]

Answer 16

and the 4 means basically nothing?

Answer 17

oh no it means something
it tells you what the foci are
but in this case it tells you how to find \(b\)

Answer 18

ok. thanks!!!!

Answer 19

yw

Answer 20

how would you solve this type of problem but for a hyperbola?