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# What is the answer to (y-3)^2/16 - (x+2)^2/9=1 in hyperbola for, with the center, vertices, foci, asymps, and what a,b, and c equal?

Please show all work so I can understand how to do it this is my last problem and I am stuck.

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### 2 Answers

I differ from the above answers on the following points----
I take up only those points where my answers are different.

Verities are ---  (-2,5).   ,  (-2,1)

Foci are  ---       (-2,8).  ,   (-2,-2)

Asymptotes ----- y= -4x/3. +1/3

y= 4x/3+17/3

First, we have to understand what the parts of the equation represent, and how it relates to the given problem.

GENERAL EQUATION OF A VERTICAL  HYPERBOLA:

(y-k)2/a2 - (x-h)2/b2 = 1

vertices: a2 = 16 -> a = ±4

b2 = 9 -> b = ±3

to find c2: c2 = a2 + b=> c2 = 16 + 9 => c2 = 25 => c = ± 5 => foci: (0,±c) => foci:(0,±5)

x-intercept: (±b,0) = (3,0),(-3,0)

center: (h,k) = in our case, is (-2,3). **Note, our x-coordinate of our center is -2 because x-(-2) = x+2**

latus rectum: 2b2/a = 2(9)/4 = 18/4 = 9/2

asymptotes: (y-3)2/16 - (x+2)2/9 = 0 => (y-3)2/16 = (x+2)2/9 => Next, cross-multiply to get 9(y-3)2/9 = 16(x+2)2/9 => (y-3)2 = 16(x+2)2/9 => take the square root of both sides => y-3 = 4(x+2)/3=> add 3 to both sides to solve for y => y = [4(x+2)/3)+3]

Ok, so, now we have all the pieces of the puzzle to draw it!