Question

What are the vertices of the ellipse given by the equation (x+3)^2/16 + (y-4)^2/121=1

Answer 1

someone help-_-

Answer 2

To figure out the center, look at the inside of the two parenthesis. We have
(x+3) and (y-4)

Answer 3

ok now what

Answer 4

What formula would I use?

Answer 5

@Hero Help !? Pleaseee

Answer 6

So the center is just whatever x and y values will make those parenthesis both 0.

Answer 7

@FlyinSolo_424 You said "OK", so what is your ( h, k) ???

Answer 8

We've got a vertical ellipse here because a (=sqrt(16)) < b (=sqrt(121)). That means the value of x that makes (x+3) = 0 will be the x value for our vertices, and we need to find the values of y. If the x term = 0, we only need to solve \[(y-4)^2/121 = 1\] which reduces to finding the values of y such that |y-4|=11 (because sqrt(121) = 11). Hopefully you agree that those values are y = 15 and y = -7. That gives us vertices at (-3,15) and (-3,-7).