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?

Answer 1

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

The x-axis is the major axis of the hyperbola since it contains both of the foci.

Since the length of the major axis 2a = 10 is the difference between the focal radii, 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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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