Question

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

Answer 1

standard form of an ellipse with vertices at h and k is given by

\[(x-h)^{2}/a ^{2} + (y-k)^{2}/b ^{2}= 1\]

Answer 2

so compare and get the answer !

Answer 3

so the vertices are h and k right??

Answer 4

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

Answer 5

it would be 1 and 3

Answer 6

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

Answer 7

so.. what is it??

Answer 8

it would be -1 and -3 ??

Answer 9

Bingo!

Answer 10

so thats the vertices??and thats it??

Answer 11

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

Answer 12

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

Answer 13

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)

Answer 14

Here is a graph of it

Answer 15

thank you both :)