Find the lengths of the axes, the coordinates of the vertices and the foci, the eccentricity and length of the latus rectum of the hyperbola
`(x^(2))/(36)-(y^(2))/(64)=1.`

Explanation

The equation of the given hyperbola is `(x^(2))/(36)-(y^(2))/(64)=1.`
Comparing the given equation with `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1,` we get
`a^(2)=36 and b^(2)=64.`
`therefore a=6,b=8and c=sqrt(a^(2)+b^(2))=sqrt(36+64)=sqrt(100)=10.`
(i) length of the transverse axis `=2a=(2xx6)` units = 12 units .
Length of the conjugate axis `=2b=(2xx8)` units =16 units.
(ii) The coordinates of the vertices are `A(-c,0) and F_(2)(c,0),i.e., F_(1) (-10,0) and F_(2)(10,0).`
(iv) Eccentricity ,`e=(c)/(a)=(10)/(6)=(5)/(3).`
(v) Length of the latus rectum `(2b^(2))/(a)=((2xx64)/(6))"units"=(64)/(3)"units".`