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

User Icon for bobpursley bobpursley answered
16 years ago

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.

User Icon for Anonymous Anonymous answered
16 years ago

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).

User Icon for shankeria shankeria answered
14 years ago

If the exterior angle of a regular polygon is 45o, then find the number of sides of the polygon

User Icon for Explain Bot Explain Bot answered
1 year ago

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).