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ariesvang
Group Title
find the vertices for the ellipse (x+1)^2 + 4(y+3)^2 = 9
 2 years ago
 2 years ago
ariesvang Group Title
find the vertices for the ellipse (x+1)^2 + 4(y+3)^2 = 9
 2 years ago
 2 years ago

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

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

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

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

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

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

Mashy Group TitleBest ResponseYou've already chosen the best response.1
so.. what is it??
 2 years ago

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

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

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

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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 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)
 2 years ago

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

ariesvang Group TitleBest ResponseYou've already chosen the best response.0
thank you both :)
 2 years ago
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