What is the answer to (y-3)^2/16 - (x+2)^2/9 in hyperbola form with the center, vertices, foci, asymps, and what a,b, and c equal?

Question

Please show work so I will understand I have been stuck on this one problem.

Answer 1

Recall the standard form for the equation of a hyperbola is as follows:

     Horizontal transverse axis:     (x - h)²/a²  -  (y - k)²/b²  = 1  

     Vertical transverse axis:     (y - k)²/a²  -  (x - h)²/b²  = 1 

The equation of the hyperbola in question is already in standard form:

          (y - 3)²/16  -  (x + 2)²/9  = 1

Looking at the equation, we see that the transverse axis is vertical. With this, we can find the center, vertices, foci, and asymptotes.

   h = -2 ,   k = 3      ==>     center:   (h, k) = (-2, 3)

   a² = 16     ==>     √a² = √16     ==>     a = 4

   b² = 9     ==>     √b² = √9     ==>     b = 3

   to find c, use the following formula:      c² = a² + b² 

                                                            c² = 16 + 9

                                                            c² = 25      ==>      c = 5

Vertices:     (h, k + a)  =  (-2, 3 + 4) = (-2, 7)

                    (h, k - a)  =  (-2, 3 - 4) = (-2, -1)

Foci:     (h, k + c)  =  (-2, 3 + 5) = (-2, 8)

             (h, k - c)  =  (-2, 3 - 5) = (-2, -2)

Asymptotes:     y - k = ± (a/b)(x - h)

                         y - 3 = ± (4/3)(x + 2)

         thus, the asymptotes of this hyperbola have the following equations:

            y - 3 = (4/3)(x + 2)          and          y - 3 = -(4/3)(x + 2)