Need help with a problem. X^2 + (y+1)^2/3 =1. It is x squared plus y+1 squared divided by 3 =1. Don't know how to start it. Please help
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I believe it's an ellipse.
Originally Posted by Nfalconer Need help with a problem. X^2 + (y+1)^2/3 =1. It is x squared plus y+1 squared divided by 3 =1. Don't know how to start it. Please help Is it $x^2+\dfrac{(y+1)^2}{3}=1$ or $\large{x^2+(y+1)^{\frac{2}{3}}=1?}$
It is the first one. I am trying to figure out foci and vertices. I know it is a vertical ellipse but that's it. Need to find a b and c.
Originally Posted by Nfalconer It is the first one. I am trying to figure out foci and vertices. I know it is a vertical ellipse but that's it. Need to find a b and c. Do you see how you could have avoided that confusion? Write as $x^2+\frac{(y+1)^2}{3}=1$. The center of the ellipse is $(0,-1)$. The foci are equally distance on each side of the center on the major axis. What is the major axis?
I got foci 1 as (0,.4) and foci 2 as ( 0, -2.4)
For the vertices I got (0,.7) and (0,-2.7)?
Originally Posted by Nfalconer I got foci 1 as (0,.4) and foci 2 as ( 0, -2.4) How did you get those values?
For foci 1 I did (h,k+c) so 0, -1+1.4=(0,.4) and foci 2 I did (h, k-c)= (0, -1-1.4)= (0,-2.4). Is that wrong?
And the vertices I did (0, -1+1.7) and (0,-1.7+-2)
Originally Posted by Nfalconer And the vertices I did (0, -1+1.7) and (0,-1.7+-2) Tell us how you got $c$. Be detailed, please.
I got c by doing 3-1=2 then got +or- 1.4
I did A(3) - B(1) to get 2
I messed up everything didn't I?
Originally Posted by Nfalconer And the vertices I did (0, -1+1.7) and (0,-1.7+-2) I'm going to slip this in here: It is better to give exact values rather than use decimals. ie. $\displaystyle ( 0, ~ -1 + \sqrt{3} ); ( 0, -1 - \sqrt{3} )$ -Dan