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

Jan 9, 2018

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= $\sqrt{\frac{{a}^{2} + {b}^{2}}{a} ^ 2}$
we get e=$\frac{\sqrt{202}}{9}$
foci will be (+/- ae,0)
which are
(+/- 11$\frac{\sqrt{202}}{9}$,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.
$g r a p h \left\{{x}^{2} / {11}^{2} - {y}^{2} / {9}^{2} = 1 \left[- 40 , 40 , - 20 , 20\right]\right\}$