1. ## Ellipses

I get the general idea of finding the Verticies, Foci, Eccentricity, Major Axes, and the Minon Axes.

However i'm having trouble doing it with this problem

2x^2 + y^2 = 3

Can someone point me in the right direction? I know I need to get it to equal 1 so I divide the entire equation by 3. This in turn will give me

(2/3)x^2 + (y^2)/ 3 = 1

Well looking at the numbers, x has a greater value than y since it is 2/3. So the verticies should be (+-Sqrt(2/3),0) but when I check the back of the book, it's not giving me that answer. It gives me (0,+-sqrt(3))

Help

2. Another one i'm having problems with is

x^2 + 4y^2 = 1

3. Originally Posted by JonathanEyoon
I get the general idea of finding the Verticies, Foci, Eccentricity, Major Axes, and the Minon Axes.

However i'm having trouble doing it with this problem

2x^2 + y^2 = 3
Yes, you need to start by dividing by 3. But you need to put it into the correct form:
$2x^2 + y^2 = 3$

$\frac{2}{3}x^2 + \frac{1}{3}y^2 = 1$

$\frac{x^2}{\frac{3}{2}} + \frac{y^2}{3} = 1$

$\frac{x^2}{\left ( \sqrt{ \frac{3}{2}} \right ) ^2} + \frac{y^2}{(\sqrt{3})^2} = 1$

So $a = \sqrt{3}$ is the semimajor axis and $b = \sqrt{\frac{3}{2}}$ is the semiminor axis.

-Dan

4. That helps me out quite a bit

Originally Posted by topsquark
Yes, you need to start by dividing by 3. But you need to put it into the correct form:
$2x^2 + y^2 = 3$

$\frac{2}{3}x^2 + \frac{1}{3}y^2 = 1$

$\frac{x^2}{\frac{3}{2}} + \frac{y^2}{3} = 1$

$\frac{x^2}{\left ( \sqrt{ \frac{3}{2}} \right ) ^2} + \frac{y^2}{(\sqrt{3})^2} = 1$

So $a = \sqrt{3}$ is the semimajor axis and $b = \sqrt{\frac{3}{2}}$ is the semiminor axis.

-Dan

5. So the second one I posted I'm having a problem with should be?

6. Originally Posted by JonathanEyoon
So the second one I posted I'm having a problem with should be?
$x^2 + 4y^2 = 1$

$\frac{x^2}{1^2} + \frac{y^2}{ \left ( \frac{1}{2} \right ) ^2} = 1$

-Dan

7. Originally Posted by topsquark
$x^2 + 4y^2 = 1$

$\frac{x^2}{1^2} + \frac{y^2}{ \left ( \frac{1}{2} \right ) ^2} = 1$

-Dan

For the denominator in Y^2, how did you get 1/2?

8. Originally Posted by JonathanEyoon
For the denominator in Y^2, how did you get 1/2?
Step by step, then.

$x^2 + 4y^2 = 1$

$\frac{x^2}{1} + \frac{y^2}{\frac{1}{4}} = 1$

$\frac{x^2}{1^2} + \frac{y^2}{\left ( \frac{1}{2} \right ) ^2} = 1$

-Dan