9x^2+16y^2-18x+64y-71=0
find the coordinates of the center, the foci, and the vertices of this ellipse.
9x^2+16y^2-18x+64y-71=0
9x^2-18x+16y^2+64y=71
9(x^2-2x)+16(y^2+4y)=71
9(x^2-2x+1)+16(y^2+4y+4)=71
9x^2 - 18x + 16y^2 + 64y = 71
9(x^2 - 2x) + 16(y^2 + 4y) = 71
9(x^2- 2x + 1) + 16(y^2 + 4y + 4) = 71 + 9 + 64
9(x - 1)^2 + 16(y + 2)^2 = 144
(x - 1)^2/16 + (y + 2)^2/9 = 1
centered at (1, -1), major axis length 8, minor axis length 6.
Vertices are at (5, -1) and (-3, -1).
The last line is just wrong. You added 9 +64 to the left, and did nothing to the right. If you add things to one side, you have to add to the other.
If the exterior angle of a regular polygon is 45o, then find the number of sides of the polygon
To find the center, foci, and vertices of the ellipse represented by the equation 9x^2+16y^2-18x+64y-71=0, we need to rearrange the equation into a standard form.
First, we complete the square for both the x and y terms.
Starting from the given equation:
9(x^2-2x)+16(y^2+4y)=71
For the x terms:
9(x^2-2x) can be rewritten as 9((x-1)^2 - 1)
For the y terms:
16(y^2+4y) can be rewritten as 16((y+2)^2 - 4)
Now we have:
9((x-1)^2 - 1) + 16((y+2)^2 - 4) = 71
Expanding the equation:
9(x-1)^2 - 9 + 16(y+2)^2 - 64 = 71
Rearranging terms:
9(x-1)^2 + 16(y+2)^2 = 144
Dividing both sides by 144:
(x-1)^2/16 + (y+2)^2/9 = 1
Now the equation is in the standard form:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
Where (h,k) represents the coordinates of the center, and a and b represent the semi-major and semi-minor axes of the ellipse.
Comparing this with our equation, we can see that h = 1, k = -2, a^2 = 16, and b^2 = 9.
Therefore, the center of the ellipse is at the coordinates (1,-2).
To find the vertices, we can utilize the fact that the distance between the center and each vertex is equal to the semi-major axis, a.
So, the coordinates of the vertices are obtained as follows:
Vertex 1: (h+a, k)
Vertex 2: (h-a, k)
Substituting the known values:
Vertex 1: (1+4, -2) = (5, -2)
Vertex 2: (1-4, -2) = (-3, -2)
Thus, the vertices of the ellipse are (5, -2) and (-3, -2).
To find the foci, we can use the Pythagorean relation between the distances from the foci to each point on the ellipse:
c^2 = a^2 - b^2
Where c represents the distance from the center to each focus.
Calculating the values:
c^2 = 16 - 9
c^2 = 7
Taking the square root of both sides:
c = √7
The coordinates of the foci are obtained as follows:
Focus 1: (h+c, k)
Focus 2: (h-c, k)
Substituting the known values:
Focus 1: (1+√7, -2)
Focus 2: (1-√7, -2)
Therefore, the foci of the ellipse are (√7+1, -2) and (-√7+1, -2).