What is the center of a hyperbola?

Answer 1

Use this as an example:

9x2-16y2+18x+160y-247=0

First put the equation into standard form.

9x² - 16y² + 18x + 160y - 247 = 0

Now complete the square.

9(x² + 2x + 1) - 16(y² - 10y + 25) = 247 + 9 - 400

9(x + 1)² - 16(y - 5)² = -144

Multiply thru by -1 since the right hand side is negative.

16(y - 5)² - 9(x + 1)² = 144

Set equal to one.

(y - 5)²/9 - (x + 1)²/16 = 1

Since y² is the positive squared term, the pair of hyperbolas open vertically up and down.

The center (h,k) = (-1,5).

a² = 9 and b² = 16

a = 3 and b = 4

The vertices are (h,k-a) and (h,k+a) or

(-1,5-3) and (-1,5+3) which is (-1,2) and (-1,8).

c² = a² + b² = 9 + 16 = 25

c = 5

The foci are (h,k-c) and (h,k+c) or

(-1,5-5) and (-1,5+5) which is (-1,0) and (-1,10).