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Question 167819: Can I get some help with this problem as well? I posted it the other day and did not get a response. Thanks in advance.
Find an equation of the hyperbola centered at the origin with vertice at
( 0, 3 ) and focus at (0,5).
Answer by midwood_trail(310)   (Show Source):
You can put this solution on YOUR website! The standard form for an equation of the hyperbola centered at the origin
is the following: (x^2/a^2) + (y^2/b^2) = 1
Now, you were given one vertices to be the point (0, 3) and one focus to be the point (0, 5). I assume that you were given vertices (0,3) and (0,-3)
and that you were also given the foci to be (0,5)and (0, -5).
From the information we are to make an equation that will look like the standard form as written above.
STEPS:
1-Find whether the transverse axis is vertical or horizontal.
In this question, the vertices and the foci lie on the y–axis. Therefore, the transverse axis is vertical.
2-The foci and the vertices are equidistant from the origin. Find the origin.
The center is the origin since the foci and the vertices are equidistant from the origin.
3-Find c and a, using the foci and the vertices.
Since the foci are each 5 units from the center, c = 5.
Similarly, the vertices are each 3 units from the center, a = 3.
4-Find b using the equation b^2 = c^2 – a^2.
You know a to be 3 and c to be 5. So, plug and chug.
b^2 = (5)^2 - (3)^2
b^2 = 25 - 9
b^2 = 16
Taking the square root of both sides we get b = 4
Now that we have the values of a and b, we plug that into the standard form given above, simplify and that will be your equation.
The standard form for an equation of the hyperbola centered at the origin
is the following: (x^2/a^2) + (y^2/b^2) = 1
Let a = 3 and b = 4
(x^2/3^2) + (y^2/4^2) = 1
Final answer: (x^2/9) + (y^2/16) = 1
Did you follow?
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