Question 133152: Find the vertices of a hyperbola with the equation (x + 3)^2 - 4(y - 2)^2 = 4.
a. (-4, 2) and (-2, 2)
b. (-5, 2) and (-1, 2)
c. (4, -3) and (0, -3)
d. (3, -3) and (1, -3)
Answer by nycsharkman(136)   (Show Source):
You can put this solution on YOUR website! Find the vertices of a hyperbola with the equation (x + 3)^2 - 4(y - 2)^2 = 4.
a. (-4, 2) and (-2, 2)
b. (-5, 2) and (-1, 2)
c. (4, -3) and (0, -3)
d. (3, -3) and (1, -3)
The first thing you want to do is to write the equation in standard form.
All hyperbola standard form equations = 1.
So, this means you must divide everthing by 4 to get 1 on the right side.
Why 4? Because your equation equals 4 and we want it to equal 1. As you know, anything divided by itself = 1.
Dividing everything by 4 and the result is:
[(x + 3)^2]/4 - [4(y - 2)^2]/4 = 1
Now this looks like the general form:
[(x - h)^2]/a^2 - [(y - k)^2]/b^2
The vertices are:
(h + a, k) and (h - a, k)
We know the value of h and k as given in the equation above.
h = -3 and k = 2...The opposite of the original signs must be used. In other words, 3 became -3 and -2 became 2 as the new values for h and k.
What is a?
Do you see the denominator 4 in the fraction [(x + 3)^2]/4 above?
If I equate 4 to a^2 and take the square root, I will get 2 as the value of a.
So, a = 2. We reject -2 because vertices represent distance and distance CANNOT be negative.
I will now plug the values of h, k and a into (h + a, k) and (h - a, k) and find the vertices.
(-3 + 2, 2) = (-1, 2)
(-3 -2, 2) = (-5, 2)
Answer: Choice B
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