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)

## im having trouble with some math problems can you help

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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}= 1With 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( a

^{2}+ b^{2}) , so 6 = sqrt( 16 + b^{2}) . 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}= 1The 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.

# Comments

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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