Question

foci(17+/,0), vertices (+/8,0)
c^2 = a^2 – b^2

c = 17 and a = 8
c^2= 289 and a^2 = 64
a^2= c^2 – b^2
64 = 289 – 225
Equation of ellipse x^2/225+y^2/289=1
Is this right?

Answer 1

@phi

Answer 2

@texaschic101

Answer 3

shouldn't the vertices be outside of the focus points ?
In other words, are you sure you didn't switch them ?

Answer 4

Probably. If it makes a difference, then what should it be?

Answer 5

either that, or you have a hyperbola ?

What information are you given?

Answer 6

It is, Write an equation of a hyperbola with the given foci and vertices.

Answer 7

then it's clear it's not an ellipse. so your answer has to be tweaked.

Answer 8

This gives the details
http://www.purplemath.com/modules/hyperbola.htm

Answer 9

Am I suppose to use a different equation?

Answer 10

Based on the picture at the posted link, and the equation they give, the equation should be
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]
we need to find a and b

a is the distance from the center to the vertex.
the center is exactly between the 2 vertexes, i.e. between (-8,0) and (+8,0) , which is (0,0)

Answer 11

so you did this part correctly
c = 17 and a = 8
c^2= 289 and a^2 = 64

to find b (for a hyperbola) the link says use this formula
\[ c^2 = a^2 + b^2 \]
or , solving for b^2,
\[ b^2 = c^2 - a^2\]

Answer 12

can you find b ?

Answer 13

50625?

Answer 14

I think that is b^2

Answer 15

then b is 225

Answer 16

no, b^4

Answer 17

b^4?

Answer 18

Am I looking for b^4 or am I writing the equation?

Answer 19

You should be finding b using
b^2 = c^2 - a^2
c=17, a=8

Answer 20

or just find b^2 (don't bother taking the square root to find b)

Answer 21

I did find b2, you said that's wrong b=225 b^2=50625, otherwise I have no idea what you're referring to.

Answer 22

b^2 is 225 (and b is 15).

Answer 23

Oh. . .I don't know where 15 came from.

Answer 24

Oh nevermind sqrt225.

Answer 25

How did you find b^2. what steps did you do?

Answer 26

289-64. . .>~>

Answer 27

and that gives 225
in other words b^2 is 225

I hope you are not getting confused about what is squared and what isn't.

But anyways, a = 8 and a^2 is 64
b is 15, and b^2 is 225
use those numbers in
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

Answer 28

You confused me by asking me what b^2 was, and I already told you. xD

Answer 29

x^2/15 - y^2/225=1?

Answer 30

is a^2 = 15 ? because you want a^2 under the x^2

Answer 31

The other way around?

Answer 32

no. First, look at what you found so far:
what is a? a is 8
what is a^2 =a*a ? it is 64
what is b? b is 15. b^2 = b*b = 225

Now look at the equation
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

replace a^2 with the number for a^2
replace the b^2 with its number

Answer 33

Ohhh, kekeke. x^2/64-y^2/225

Answer 34

almost. equations have = signs in them

Answer 35

=1

Answer 36

Here is your hyperbola

Answer 37

. . .Eh, that sucks. Should have used the calculator. Thanks as always.

Answer 38

Working with the numbers "by hand" is the only way to get reasonably good at it. And though it might not seem so, using numbers is a good thing to be able to do.

Answer 39

Yup, if phi says so. ^^