Question

Someone PLEASE help? I don't get these problem!

1.) find the standard form or the equation of the ellipse with vertices (±9,0) and eccentricity e=1/3.

2.) Find the standard form of the equation of the hyperbola with the given characteristics.
vertices: (2, - 7), (2, - 1),
foci: (2, - 9), (2, 1)

3.) Find the standard form of the equation of the ellipse with the given characteristics.
center: (-1, 7)
a = 3c
foci: (- 2, 7), (0, 7

Answer 1

Well does your course cover conics? If it does, then you should be familiar with the standard equations of ellipse and hyperbola, and then its a simple matter of substitution

Answer 2

center is half way beteen \((-9,0)\) and \((9,0)\) so the center is \((0,0)\) which makes things easier

Answer 3

equation will look like \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and what you need now is \(a\) and \(b\)

Answer 4

so this isnt an x-h/a+y-k/b?

Answer 5

oh wait now i see. sorry

Answer 6

no since \(h=k=0\) in this case

Answer 7

now you have \(a=9\) because the vertices are \((-9,0)\) and \((9,0)\)

Answer 8

\[\frac{x^2}{81}+\frac{y^2}{b^2}=1\]

Answer 9

now you need \(b\)

Answer 10

is b 9?

Answer 11

idk we have to find it, but i doubt it

Answer 12

ok how do you find b?

Answer 13

you have the eccentricity as \(\frac{1}{3}\) right?

Answer 14

yeah its 1/3

Answer 15

and we also know \(a=9\)
in general the eccentricity is \(\frac{c}{a}\) where \(c=\sqrt{a^2-b^2}\) or \(c^2=a^2-b^2\)

Answer 16

so we can solve
\[\frac{1}{3}=\frac{81-b^2}{9}\] for \(b\) i guess

Answer 17

ok that was wrong, i mean solve
\[\frac{1}{3}=\frac{\sqrt{81-b^2}}{9}\]

Answer 18

i got ±9

Answer 19

for b

Answer 20

no i don't think so
lets solve

Answer 21

\[\frac{1}{3}=\frac{\sqrt{81-b^2}}{9}\] mulitply both sides by 9 to get
\[3=\sqrt{81-b^2}\] square and get
\[9=81-b^2\] so \[b^2=81-9=72\] making \(b^2=72\)

Answer 22

final answer i think is
\[\frac{x^2}{81}+\frac{y^2}{72}=1\] lets check it

Answer 23

yes, it is right, here is the check
note how i wrote "ellipse" first so it gives all the details
http://www.wolframalpha.com/input/?i=ellipse+x^2%2F81%2By^2%2F72%3D1

Answer 24

thank you so much!

Answer 25

yw
!

Answer 26

So i know how to get the center fine. but is it the same thing for the foci? to get a and b?

Answer 27

@satellite73

Answer 28

the foci you get from \(c\)

Answer 29

and \(c\) you get from \(a\) and \(b\)
it is \(c^2=a^2-b^2\) or \(c=\sqrt{a^2-b^2}\)

Answer 30

thats what i want to know, how to get a and b. thats always the part i get wrong

Answer 31

really it depends on what you know
in the first one we knew \(a=9\) and \[\frac{1}{3}=\frac{c}{a}=\frac{c}{9}\] we used that to find \(b^2\)

Answer 32

oh ok, i see now. i thought they were different for ellipses and hyperbolas

Answer 33

Find the standard form of the equation of the ellipse with the given characteristics. center: (-1, 7) a = 3c foci: (- 2, 7), (0, 7)\[\frac{(x+2)^2}{a^2}+\frac{(y-7)^2}{b^2}=1\]

Answer 34

oops \[\frac{(x+1)^2}{a^2}+\frac{(y-7)^2}{b^2}=1\]

Answer 35

now here the foci are 1 unit to the left and right of the center
this tells you \(c=1\)

Answer 36

and since you are told that \(a=3c\) given that \(c=1\) this gives you \(a=3\) and so \(a^2=9\)

Answer 37

\[\frac{(x+2)^2}{9}+\frac{(y-7)^2}{b^2}=1\]

Answer 38

now you need \(b^2\) and again \(a^2=b^2+c^2\) so
\[9=b^2+1\] making
\[b^2=8\]

Answer 39

ok i think i got that, thanks!

Answer 40

\[\frac{(x+1)^2}{9}+\frac{(y-7)^2}{8}=1\]
http://www.wolframalpha.com/input/?i=ellipse+ \frac{%28x%2B1%29^2}{9}%2B\frac{%28y-7%29^2}{8}%3D1

Answer 41

if you want to read a really good clear explanation of all of these, with good worked out examples, try this
http://www.purplemath.com/modules/ellipse.htm

Answer 42

ok thanks again

Answer 43

or even this
http://www.purplemath.com/modules/conics.htm
yw