00:01
We're going to take the equation and convert this to a standard form of an equation of an ellipse so that we can find the center, both by vertices by thinking about the a, b, and c values, and we'll produce the graph.
00:17
The first thing i'm going to do is rewrite this, and we're going to think about completing squares here.
00:33
And we're going to subtract the four.
00:37
Now i'm also going to factor out a 4 from just the y's, which would take that negative 8 down to a negative 2.
01:05
And we'll do this very carefully here.
01:08
So to complete the square with just the xes, we take half of the 4, which is 2 and square it to get 4, so that we can write it as x plus 2 squared a perfect square.
01:21
But we have to add 4 to the right side of the equation as well to stay balanced.
01:28
For the y's, if we take half of negative 2, we get negative 1 square, we get 1.
01:35
So it would be 4 times y minus 1 squared.
01:43
But be careful, we're going to have to not just add 1 to the right hand side, 4 times 1 so add another 4 so we had a negative 4 plus 4 plus 4 we are equal to 4 now we'll divide everything by 4 so that we are an equation that is equal to 1 so that we are in standard form so we'd have y minus 1 squared over 1 which we don't need to write that but i will and xx plus 2 squared over 4.
02:30
Okay.
02:31
Now we're going to work off of this form here because this allows us to see that the center should be at negative 2 positive 1.
02:44
That the a value must be, if this is the a squared value, then a must be 2.
02:54
1 squared is 1, or the square to 1 is 1, so the b value is 1...