00:01
Hello everyone, in this problem we are given with the equation 16x square plus y square minus 32x minus 6y plus 9 which is equal to zero which is an ellipse equation.
00:18
Now we need to find its center vertices and the force.
00:22
So now we know that the ellipse equation will be in the form x minus h the whole square divided by b square plus y minus k the whole square divided by a square which is equal to 1.
00:40
So now we need to convert the given equation in this standard form.
00:46
So we are given with the equation 16x square plus y square minus 32x minus 6y plus 9 which is equal to zero.
00:58
So now we can rewrite this equation by factoring out the coefficient of the square term.
01:04
So it will be 16 of x square minus 2x plus y square minus 6y to be equal to minus 9.
01:19
So now dividing the coefficient of the square term by 16 so we will be getting it as x square minus 2x divided by 1 plus y square minus 6y divided by 16 which is equal to minus 9 by 16.
01:41
Now converting this square term these terms to this whole square term so now adding plus 1 on both sides and for here adding plus 9 by 16 on both sides so we get x square minus 2x plus 1 by 1 plus y square minus 6y plus 9 by 16 which will be equal to minus 9 by 16 plus 1 plus 9 by 16.
02:23
So simplifying here and also here it will become x minus 2 the whole square divided by 1 square plus y minus 3 the whole square divided by 4 square which is equal to 1.
02:41
So now we have obtained the recovered standard form of the given let's equation.
02:52
So now comparing the obtained equation with the standard form we have the values of h as 1, k to be 3, a to be 4 and b to be 1.
03:05
So with these values we can calculate the values of the center for the unvertical center will be given as hk and it will be substituting its values it will be 1 3.
03:21
So the required center of the ellipse is 1 3.
03:25
Now let us find the vertices...