Find the center, foci, and vertices of this hyperbola 16x^2-y^2-32x+8y+16=0

16x^2 - 32x + 16 - y^2 + 8y - 16 = -16 + 16 + 16

16(x-1)^2 - (y-4)^2 = 16

(x-1)^2 - (y-4)^2/16 = 1

h=1, k=4
a=1, b=4, c=√17
center: (1,4)
foci: (1-√17,4) (1+√17,4)
vertices: (0,4) (2,4)

To find the center, foci, and vertices of the hyperbola, we need to rewrite the equation of the hyperbola in the standard form:

(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1

where (h, k) is the center of the hyperbola, and a and b are the lengths of the transverse and conjugate axes.

First, let's rearrange the given equation:

16x^2 - 32x - y^2 + 8y + 16 = 0

Complete the square for x terms by adding (32/16)^2 = 4 to both sides:

16x^2 - 32x + 4 - y^2 + 8y + 16 = 4

Rearrange the equation:

16x^2 - 32x - y^2 + 8y = -16

Complete the square for x terms:

16(x^2 - 2x) - (y^2 - 8y) = -16

Factor out the common coefficients:

16(x^2 - 2x + 1) - (y^2 - 8y + 16) = -16

Simplify further:

16(x - 1)^2 - (y - 4)^2 = -16

Divide everything by -16 to isolate the equation on the right side:

(x - 1)^2 / (-1) - (y - 4)^2 / 16 = 1

Now we can determine the center, foci, and vertices:

The center of the hyperbola is given by (h, k), which in this case is (1, 4).

To find the distance of the transverse axis, we take the square root of the denominator of x. So, a = sqrt(1) = 1.

To find the distance of the conjugate axis, we take the square root of the denominator of y. So, b = sqrt(16) = 4.

The vertices are located at a units to the left and right of the center, so the vertices are (1 - a, 4) and (1 + a, 4), which in this case are (1 - 1, 4) and (1 + 1, 4), giving us the vertices (0, 4) and (2, 4).

The foci are located at a units to the left and right of the center, so the foci are (1 - a, 4) and (1 + a, 4), which in this case are (1 - 1, 4) and (1 + 1, 4), giving us the foci (0, 4) and (2, 4).

Therefore, the center of the hyperbola is (1, 4), the vertices are (0, 4) and (2, 4), and the foci are (0, 4) and (2, 4).

To find the center, foci, and vertices of the given hyperbola, we need to rewrite the equation in standard form. The standard form for a hyperbola centered at (h, k) is:

((x - h)^2) / a^2 - ((y - k)^2) / b^2 = 1, if the hyperbola opens horizontally
((y - k)^2) / a^2 - ((x - h)^2) / b^2 = 1, if the hyperbola opens vertically

The given equation is: 16x^2 - y^2 - 32x + 8y + 16 = 0

Rearranging the equation, we group the x and y terms separately and complete the square for each variable:

(16x^2 - 32x) - (y^2 - 8y) = -16
16(x^2 - 2x) - (y^2 - 8y) = -16

Completing the square for x:
16(x^2 - 2x + 1) - (y^2 - 8y) = -16 + 16
16(x - 1)^2 - (y^2 - 8y) = 0

Completing the square for y:
16(x - 1)^2 - (y^2 - 8y + 16) = 0 + 16
16(x - 1)^2 - (y - 4)^2 = 16

Dividing both sides by 16 to isolate the equation on the right-hand side:
((x - 1)^2) / 1 - ((y - 4)^2) / 16 = 1

Comparing the obtained equation with the standard form, we can determine the values of a, b, h, and k.

a = 1
b = 4
h = 1
k = 4

The center of the hyperbola is (h, k) = (1, 4).

To find the foci, we can use the formula c^2 = a^2 + b^2, where c represents the distance from the center to each focus.

c^2 = a^2 + b^2
c^2 = 1^2 + 4^2
c^2 = 1 + 16
c^2 = 17

Taking the square root of both sides:
c = √17

The foci are located on the x-axis, and their coordinates are (h ± c, k). Therefore, the coordinates of the foci are:

F1 = (1 + √17, 4)
F2 = (1 - √17, 4)

Finally, to find the vertices of the hyperbola, we can use the following formulas:

For a hyperbola opening horizontally: (h ± a, k)
For a hyperbola opening vertically: (h, k ± a)

The vertices are the points that are a distance of a units away from the center along the major axis. Therefore, the coordinates of the vertices are:

V1 = (1 + 1, 4) = (2, 4)
V2 = (1 - 1, 4) = (0, 4)

Hence, the center, foci, and vertices of the given hyperbola 16x^2 - y^2 - 32x + 8y + 16 = 0 are as follows:

Center: (1, 4)
Foci: (1 + √17, 4) and (1 - √17, 4)
Vertices: (2, 4) and (0, 4)