How do you find all the critical points to graph #x^2/9 -y^2/16=1# including vertices, foci and asymptotes?

1 Answer
Jun 10, 2016

Answer:

Focii are #(+-a*e,0) = (+-5,0)#
Vertices are #(+-a,0)= (+-3,0), (0,+-b)=(0,+-4)#
Eqns. of asymptotes are #x/a=+-y/b#, i.e.,#x/3=+-y/4#

Explanation:

We compare the given equation (eqn.) with the standard eqn., #x^2/a^2-y^2/b^2=1# #:.# #a^2=9, b^2=16.#
The eccentricity e is given by : # b^2-a^2(e^2-1).#
#:. 16=9(e^2-1)##:. e=5/3#
Focii are #(+-a*e,0) = (+-5,0)#
Vertices are #(+-a,0)= (+-3,0), (0,+-b)=(0,+-4)#
Eqns. of asymptotes are #x/a=+-y/b#, i.e.,#x/3=+-y/4#