1. ## eccentrity of hyperbola

find the eccentricity of the hyperbola.

x^2-2y^2=6

I got 3sqrt of 6/ 6

2. Originally Posted by scottydoint
find the eccentricity of the hyperbola.

x^2-2y^2=6

I got 3sqrt of 6/ 6
For the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ the eccentricity is given by $\displaystyle e = \sqrt{1 + \frac{b^2}{a^2}}$.

For your hyperbola $\displaystyle a = \sqrt{6}$ and $\displaystyle b = \sqrt{3}$. Therefore $\displaystyle e = \frac{\sqrt{6}}{2}$ which is what your answer simplifies to.

So yes you are correct.

3. ## two graphs

Sorry for all of these questions, but this is last minute and my teacher can't even get these problems right in class so I am trying to teach myself from a book that is hard to follow.

How do you graph this parabola:
x=-3y^2-6y+3

and how do you graph this circle? I don't think its possible.

x^2+(y-1)^2+1=0

4. Originally Posted by scottydoint
Sorry for all of these questions, but this is last minute and my teacher can't even get these problems right in class so I am trying to teach myself from a book that is hard to follow.

How do you graph this parabola:
x=-3y^2-6y+3

Mr F says: The standard form is found by completing the square:

$\displaystyle {\color{red}x = -3 (y^2 + 2y - 1) = -3 [(y+1)^2 - 2] = -3(y+1)^2 + 6} \, ....$

and how do you graph this circle? I don't think its possible.

x^2+(y-1)^2+1=0 Mr F says: $\displaystyle {\color{red} \Rightarrow x^2 + (y-1)^2 = -1}$ which is clearly not possible for real values of x and y.

5. Originally Posted by mr fantastic