SOLUTION: Hi, I have been staring at this problem for about an hour . could you please help me ?
It's (x+2)^2/9 + (y+2)^2/4 = 1
I am suppose to sketch the ellipse which I will be able
Algebra ->
Quadratic-relations-and-conic-sections
-> SOLUTION: Hi, I have been staring at this problem for about an hour . could you please help me ?
It's (x+2)^2/9 + (y+2)^2/4 = 1
I am suppose to sketch the ellipse which I will be able
Log On
Question 544552: Hi, I have been staring at this problem for about an hour . could you please help me ?
It's (x+2)^2/9 + (y+2)^2/4 = 1
I am suppose to sketch the ellipse which I will be able to do on my own, but I just don't know how to find the numbers..
If you could help me, I would appreciate it so much !
Thaaaaank you :) Answer by KMST(5328) (Show Source):
You can put this solution on YOUR website!
Think of it this way: represents a circle centered at (0,0) with radius 1.
(You can see that the formula says that the distance to the center is 1). would represent a circle centered at (-2,-2) with radius 1.
Stretching the drawing in the x and y directions to the same extent you would get a bigger circle with the same center: <--> <--> would represent a circle centered at (-2,-2) with radius 3. <--> <--> would represent a circle centered at (-2,-2) with radius 2.
Your ellipse is a stretched circle, symmetrical about axes and . It's only that the guy stretching in the x direction was pulling harder. (substitute and see) at and .
It crosses axis at and .