How do you write a standard form equation for the hyperbola with foci are (-6,0) and (6,0) and the difference of the focal radii is 10?

1 Answer
A. S. Adikesavan
Apr 23, 2016

x^2/5^2-y^2/((sqrt 11)^2)=1 or 11x^2-25y^2-275=0

Explanation:

As both the foci are on the x-axis, x-axis is the major axis of the hyperbola.

The difference between the focal radii = length of the major axis 2a = 10. So, a = 5.

The distance between the foci s(6, 0) and S'#(-6, 0) = 12 = major axis length 2 a x eccentricity = 2 a e = 10 e =12. So, e = 6/5.

The semi-transverse axis b =a sqrt(e^2-1)=sqrt 11

The standard form of the equation is x^2/a^2-y^2/b^2=1
here, it is
x^2/5^2-y^2/((sqrt 11)^2)=1 or 11x^2-25y^2-275=0.

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