PLEASE HELP W/ MY TRIG HW: Find an equation of the hyperbola *BEST ANSWER AWARDED*?

Let the equation be

[(x-0)^2]/a^2-[(y-1)^2]/b^2=1

where a, b are constants.

The focus=C(10,1) & the vertex=V(3,1)

=>10^2=3^2+b^2

=>b^2=100-9

=>b^2=91

Thus the equation of the hyperbola is

(x^2)/9-[(y-1)^2]/91=1

Center is (0,1) and hyperbola is oriented to open left/right. Your x-term will be positive since it opens left/right. So far, your numerators are x^2 and (y-1)^2. The distance squared from the center to either vertex is the denominator of the positive fraction. Distance is either -3 or 3 and squared becomes 9. Positive fraction is now x^(2)/9. Denominator of other fraction is the missing b^2 of the Pythagorean Relation of a^2 (which is 9) + b^2 = c^2 (which is the distance of the center to either focus, in this case 10^2 = 100).

So far, you have x^(2)/9 - (y-1)^(2)/b^2 = 1

a^(2) + b^(2) = c^(2)

9 + b^(2) = 100

b^(2) = 91

Answer: x^(2)/9 - (y-1)^(2)/91 = 1

(x - h)^2 / a^2 - (y - k)^2 / b^2 = 1

The center of the hyperbola is half-way between the vertices at (0,1); therefore h = 0 and k =1.

Half the length of the major axis is 3 (the distance from the center of the hyperbola to a vertex) = a; a^2 = 9.

c, the focal length is 10; so c^2 = 100.

b^2 = c^2 - a^2 = 100 - 9 = 91.

Answer: (x)^2 / 9 - (y - 1)^2 / 91 = 1