Question

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

Answer 1

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#