FIND THE CENTER AND VERTICES OF THE HYPERBOLA 27x29y254x90y4

Solution

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).