Question

How do you identify the vertices, foci, and direction of `x^2/121-y^2/81=1`?

Answer 1

Taking the general equation of hyperbola and comparing with
we get that `a^2`=121 and a =11
and `b^2`=81 and b=9.
Eccentricity(e) of the hyperbola will be
e= `sqrt((a^2 + b^2)/a^2)`
we get e=`sqrt(202)/9`
foci will be (+/- ae,0)
which are
(+/- 11`sqrt(202)/9`,0)
vertices are(+/- a,0)
which are(11,0)
as the rhs=1 the hyperbola will be towards +ve and -ve x axis with its axis as the x and y axis.
`graph{x^2/11^2 -y^2/9^2=1 [-40, 40, -20, 20]}`