If the equation represents a hyperbola, find the center, foci, and asymptotes. If the equation represents an ellipse, find the center, vertices, and foci. If the equation represents a circle, find the center and radius. If the equation represents a parabola, find the focus and directrix.
-$y ^ { 2 } - x ^ { 2 } - 14 y + 12 x - 36 = 0$
A) C: (6, 7); F: $( 0,7 + 7 \sqrt { 2 } ) , ( 0,7 - 7 \sqrt { 2 } ) ; A : y - 7 = x - 6 , y - 7 = - ( x - 6 )$
B) C: $( 6,0 ) ; F : ( 7 \sqrt { 2 } , 0 ) , ( - 7 \sqrt { 2 } , 0 ) ; A : y = \frac { 1 } { 7 } x , y = - \frac { 1 } { 7 } x$
C) C: $( 6,7 ) ; F : ( 6,7 + 7 \sqrt { 2 } ) , ( 6,7 - 7 \sqrt { 2 } ) ; A : y - 7 = x - 6 , y - 7 = - ( x - 6 )$
D) C: $( - 1,7 ) ; F : ( 7 \sqrt { 2 } , 0 ) , ( - 7 \sqrt { 2 } , 0 ) ; A : y = x , y = - x$

Question

Quizplus · Accepted Answer

First, rearrange the equation into the standard form of a hyperbola by completing the squares: $y^2 - 14y - x^2 + 12x = 36$. Completing the squares gives $(y-7)^2 - (x-6)^2 = 49$, which is a hyperbola. The center is at (6,7), the foci are calculated using $c^2 = a^2 + b^2$, where $c = 7\sqrt{2}$, so the foci are $(6, 7 + 7\sqrt{2})$ and $(6, 7 - 7\sqrt{2})$. The asymptotes are lines through the center with slopes equal to ±b/a, which gives the equations $y - 7 = ±(x - 6)$.

If the Equation Represents a Hyperbola, Find the Center, Foci y2−x2−14y+12x−36=0y ^ { 2 } - x ^ { 2 } - 14 y + 12 x - 36 = 0y2−x2−14y+12x−36=0

If the Equation Represents a Hyperbola, Find the Center, Foci $y ^ { 2 } - x ^ { 2 } - 14 y + 12 x - 36 = 0$