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Mashy
 2 years ago
Best ResponseYou've already chosen the best response.1standard form of an ellipse with vertices at h and k is given by \[(xh)^{2}/a ^{2} + (yk)^{2}/b ^{2}= 1\]

Mashy
 2 years ago
Best ResponseYou've already chosen the best response.1so compare and get the answer !

ariesvang
 2 years ago
Best ResponseYou've already chosen the best response.0so the vertices are h and k right??

Mashy
 2 years ago
Best ResponseYou've already chosen the best response.1Yes in that given general equation its, h, k.. so in your equation what is h and k?

Mashy
 2 years ago
Best ResponseYou've already chosen the best response.1No.. check properly.. if it was 1 and 3.. the equation to the ellipse would be (x1)^2 + 4(y3)^2 = 9

ariesvang
 2 years ago
Best ResponseYou've already chosen the best response.0it would be 1 and 3 ??

ariesvang
 2 years ago
Best ResponseYou've already chosen the best response.0so thats the vertices??and thats it??

Mashy
 2 years ago
Best ResponseYou've already chosen the best response.1Yup that's it.. kind of a puny question eh?? Go ask your teachers to give you better problems to solve :P

phi
 2 years ago
Best ResponseYou've already chosen the best response.1You 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

phi
 2 years ago
Best ResponseYou've already chosen the best response.1Your 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 semimajor 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 semimajor axis. x= 1+3= 2 or x= 13= 4 and the vertices are (4,3) and (2,3)

eliassaab
 2 years ago
Best ResponseYou've already chosen the best response.0Here is a graph of it
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