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im having trouble with some math problems can you help

the question is write the equation of the hyperbola with the given foci and vertices.  foci:(6,0), (-6,0)  vertices: (4,0)(-4,0)


Hi Sam;
Are these the vertices or the x-intercepts?
A parabola can only have one vertex.
It can have two x-intercepts.
Vivian, it's a hyperbola, not a parabola.

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Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...
4.9 4.9 (628 lesson ratings) (628)
This is an East - West hyperbola centered at the origin.    The standard form is (x/a)2 - (y/b)2 = 1
With this standard form, the distance from the center to a vertex is a , so   a = 4.    The distance from the center to a focal point is:
sqrt( a2 + b2)   ,  so    6 =  sqrt( 16 + b2)  .    Solving this for b  results in b =  sqrt(20).
Plugging these values for a and b into the standard form results in
(x/4)2 - (y/sqrt(20))2  = 1
The only practical way to deal with hyperbola or ellipse problems like this one is to memorize the formulas for  standard forms and distances to vertex, foci and directrix.


Hi Richard;
Thank you.  I forgot about East-West parabolas.
"memorize the formulas" -- ouch! The two foci and the four points (±a,±b) all lie on a circle centered at the origin with radius c. The two asymptotes are 1) a line through (-a,-b) and (a,b) and 2) a line through (-a,b) and (a,-b). Then you could translate by <h,k>.