Analytic geometry question: Finding the Parameters of an Ellipse

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wat2000
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Find the vertices, foci, and eccentricity of the ellipse.

1/2x^2 + 1/8y^2 = 1/4

i'm a little confused on how to clear the fraction and set the equation up to make it equal 1.
I tried to divide everything by 4 to clear it but 8 can't divide into 4 and if I use a higher number than 4 the equation won't equal 1. when the fractions are cleared the equation should be in the form of x^2/a^2 + y^2/b^2 = 1 if the equation is horizontal or x^2/b^2 + y^2/a^2 = 1 if it is vertical.
Can someone help me clear the fractions?
 
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The right hand side is equal to 1/4. So, in order to make it equal 1, you would have to multiply it by 4, not divide. Of course, anything you do to one side of the equation, you must do to the other side. Therefore, you end up with

4(1/2)x2 + 4(1/8)y2 = 4/4

Can you take it from here?
 
I have x^2/2 + y^2/1/2 = 1. I get how you told me to this but I am still a little confused on why you multiplied the left side by 4 and then divided the right side to make it equal 1. why wouldn't you multiply the right side like the left and get 16? (I know your way is right I am just trying to fully get it)
 
wat2000 said:
I have x^2/2 + y^2/1/2 = 1. I get how you told me to this but I am still a little confused on why you multiplied the left side by 4 and then divided the right side to make it equal 1. why wouldn't you multiply the right side like the left and get 16? (I know your way is right I am just trying to fully get it)

Both sides are being multiplied by 4. For the right-hand side:

(1/4) * 4 = 1

I don't know where you get 16 from.

EDIT: oh, and by the way, your left side is wrong as well. Look at the coefficient of the x2 term:

4*(1/2) = 4/2 = ?

As for the coefficient of the y2 term:

4*(1/8) = 4/8 = ?

You fill in the question marks.
 
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doesnt 4/8 = .5 or 1/2?
 
wat2000 said:
Find the vertices, foci, and eccentricity of the ellipse.

1/2x^2 + 1/8y^2 = 1/4
Can you please use parentheses? One could interpret the above as
[tex]\frac{1}{2x^2} + \frac{1}{8y^2} = \frac{1}{4}[/tex]
... which would be wrong.

wat2000 said:
doesnt 4/8 = .5 or 1/2?
So now you have
[tex]\frac{x^2}{2} + \frac{y^2}{8} = \frac{1}{4}[/tex]
[tex]\rightarrow 2x^2 + \frac{y^2}{2} = 1[/tex]

Now, rewrite the first term as a fraction.
[tex]2x^2 = \frac{x^2}{\text{?}}[/tex]
 
wat2000 said:
doesnt 4/8 = .5 or 1/2?

Yes, so the coefficient multiplying the y2 term is 1/2 i.e.:

(1/2)y2

That's not what you wrote though. You had y2 divided by 1/2 (at least, I think so, although there is some ambiguity in what you wrote due to the lack of parentheses).

Anyways, so as eumyang said, you now have:

2x2 + (1/2)y2 = 1

and you need to write these coefficients in the form (1/a2) and (1/b2). I.e.

2 = (1/a2)

a2 = ?

and,


1/2 = (1/b2)

b2 = ?