Find the foci of the hyperbola (1/10)(x-1)^2-(y-1)^2=1?

Answer 1

(x - 1)²/10 - (y - 1)²/1 = 1

So the parabola opens left and right, and a = √10 and b = 1 and the center is (1, 1).

So c² = a² + b² = 10 + 1 = 11, and the foci will be at (1 + √11, 1) and (1 - √11, 1).

See http://en.wikipedia.org/wiki/Hyperbola for more information about the formulas.

Answer 2

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

(((x - 1)^2)/10) - (((y - 1)^2)/1) = 1

a^2 = 10
b^2 = 1

10 + 1 = c^2
c^2 = 11
c = sqrt(11)

Foci = (h + c,k)(h - c,k)
Foci = (1 + sqrt(11),1)(1 - sqrt(11),1)

ANS : Foci = (1 + sqrt(11),1)(1 - sqrt(11),1)