If `(x^(2))/(36) - (y^(2))/(64) =1` represents a hyperbola, then find the co-ordinates of the foci and the vertices, the eccentricity and the length of the latus rectum.

Explanation

On comparing the given equation `(x^(2))/(36)-(y^(2))/(64) =1` with the standard of the hyperbola i.e., `x^(2))/(a^(2))-(y^(2))/(b^(2)) =1` , we get, a = 6 and b = 8 `rArr c = sqrt(a^(2)+b^(2)) = 10`
`therefore` The co-ordinates of the foci are `(pm 10, 0)` and that of the vertices are `(pm6, 0)`
The eccentricity is given by ` e = (c )/(a)`
Length of the lactus rectum ` = (2b^(2))/(a) = (2(64))/(6) = (64)/(3)`