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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)

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Hi Sam;
(4,0)(-4,0)
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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1 Answer

Hello,
 
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.

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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>.

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