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

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standard form of an ellipse with vertices at h and k is given by \[(x-h)^{2}/a ^{2} + (y-k)^{2}/b ^{2}= 1\]
so compare and get the answer !
so the vertices are h and k right??

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Other answers:

Yes in that given general equation its, h, k.. so in your equation what is h and k?
it would be 1 and 3
No.. check properly.. if it was 1 and 3.. the equation to the ellipse would be (x-1)^2 + 4(y-3)^2 = 9
so.. what is it??
it would be -1 and -3 ??
so thats the vertices??and thats it??
Yup that's it.. kind of a puny question eh?? Go ask your teachers to give you better problems to solve :P
  • phi
You should check your definitions (google works) See for what the vertex is. You found the center of the ellipse
  • 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)
Here is a graph of it
1 Attachment
thank you both :)

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