SOLUTION: How do i convert this hyperbola into standard form with these verticies (-6,0),(6,0) and focus of (-8,0),(8,0) Also,how do i convert this to standard form with these verticies, (0

All hyperbolas that have equation



look like this:



You won't have any trouble with hyperbolas if you learn all the parts
and what lengths "a", "b", and "c" stand for:

The RED CURVE with two non-connecting parts is the HYPERBOLA
The two slanted lines are the ASYMPTOTES
The green RECTANGLE is the DEFINING RECTANGLE
The VERTICES are the points V and V'
The FOCI (FOCAL POINTS) are F and F'
The CENTER is the point O. In your case the center is the origin (0,0)
but later you'll have some that have other centers.
The COVERTICES are P and Q
The TRANSVERSE AXIS is the line segment VV'. It's length is 2a
The CONJUGATE AXIS is the line segment PQ.  Its length is 2b
The SEMI-TRANSVERSE AXIS is either of the lines OV or OV'. It's length is a
The SEMI-CONJUGATE AXIS is either of the lines OA or OB. It's length is b

The FOCI (FOCAL POINTS) is "c" units from the center, that is,
the line segment OF is c units long.  

a, b, and c follow the Pythagorean theorem equation c² = a²+b²

Go here for some hyperbola questions. Be sure to click the related questions:

http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Quadratic-relations-and-conic-sections.faq.question.80329.html


Edwin