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Question 1136852: Hi. I am having a very tough time with the conic sections portion of my algebra class. Is there anyway someone can help me solve the equation:
Find the center, transverse axis, vertices, and foci of the hyperbola: y^2/81 - x^2/64 = 1
any help would be sincerely grateful!
Thank you in advance for any help!
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52797)   (Show Source):
You can put this solution on YOUR website! .
To get basic knowledge about hyperbolas, their canonical equations and all their elements, consult with my lesson
- Hyperbola definition, canonical equation, characteristic points and elements
in this site.
I guarantee you, that nowhere else you will find more clear and compact lesson on the subject.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic
"Conic sections: Hyperbolas. Definition, major elements and properties. Solved problems".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.
It is possible, that other tutors will come and present you the full solution.
I consciously don't want to do it - I think it will be 1000 times better for you to do it on your own after reading my lesson.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
(1) Determine the center.
The standard form of the equation is
In this example, since the numerators are y^2 and x^2, h and k are both 0; so the center of the hyperbola is the origin, (0,0).
(2) Determine whether the branches open up and down, or right and left.
In my experience, most students (and most teachers!) prefer to use a rule -- something like "the y^2 term is first, so the branches open in the y direction".
I prefer to use my UNDERSTANDING of what is going on, rather than a rule which might be easy to forget. Look what happens if you try to choose y=0. You end up with -x^2/64 = 1, which is not possible for real numbers. That means there is no point on the graph where y=0; and that means the branches open up and down.
(3) Transverse axis and vertices
The transverse axis is the line that passes through the points on the two branches of the hyperbola that are closest to each other. With center at the origin and the branches opening up and down, the transverse axis is the y-axis -- i.e., the line x = 0.
To find the vertices -- the points on the transverse axis that are points on the hyperbola -- set x=0 in the equation, giving you y = 9 or -9. The two vertices are (0,9) and (0,-9).
(4) Coordinates of the foci
The distance from the center of the hyperbola to each focus is c, where c^2 = a^2+b^2. In this hyperbola, that is c^2 = 81+64 = 145, so c = sqrt(145). Since the branches of the hyperbola open up and down, the coordinates of the foci are (0,sqrt(145) and (0,-sqrt(145).
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