## anonymous 4 years ago find the vertices for the ellipse (x+1)^2 + 4(y+3)^2 = 9

1. anonymous

standard form of an ellipse with vertices at h and k is given by $(x-h)^{2}/a ^{2} + (y-k)^{2}/b ^{2}= 1$

2. anonymous

so compare and get the answer !

3. anonymous

so the vertices are h and k right??

4. anonymous

Yes in that given general equation its, h, k.. so in your equation what is h and k?

5. anonymous

it would be 1 and 3

6. anonymous

No.. check properly.. if it was 1 and 3.. the equation to the ellipse would be (x-1)^2 + 4(y-3)^2 = 9

7. anonymous

so.. what is it??

8. anonymous

it would be -1 and -3 ??

9. anonymous

Bingo!

10. anonymous

so thats the vertices??and thats it??

11. anonymous

Yup that's it.. kind of a puny question eh?? Go ask your teachers to give you better problems to solve :P

12. phi

You should check your definitions (google works) See http://www.mathwords.com/v/vertices_of_an_ellipse.htm for what the vertex is. You found the center of the ellipse

13. phi

Your equation is (x+1)^2 + 4(y+3)^2 = 9 in standard form this is $\frac{(x+1)^2}{3^2}+\frac{(y+3)^2}{(\frac{3}{2})^2}=1$ The semi-major axis is the longest axis. Here it is 3 (denominator of x term) the center is (-1,-3) and when y= -3 you are on the semi-major axis. x= -1+3= 2 or x= -1-3= -4 and the vertices are (-4,-3) and (2,-3)

14. anonymous

Here is a graph of it

15. anonymous

thank you both :)