# Find the foci of the hyperbola? `x^2/64-y^2/36=1`

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If general equation of hyperbola is

`x^2/a^2+y^2/b^2=1`

then we can calculate linear eccentricity `e>0` by using formula

`e^2=a^2+b^2` **(1)**

and then the foci are `F_1(-e,0)` and `F_2(e,0)`.

Let's now apply that to your case. Here we have `a^2=64` and `b^2=36` hence

`e^2=64+36=100`

`e=10`

**So the foci are** `F_1(-10,0)` **and** `F_2(10,0).`