How do you find all the critical points to graph #x^2/9 + y^2/25 = 1# including vertices, foci and asymptotes?

1 Answer
Jun 15, 2016

Answer:

Vertices are #(-5,0)# and #(5,0)# and focii are #(0,-4)# and #(0,4)#.

Explanation:

The given equation, #x^2/9+y^2/25=1# or #x^2/3^2+y^2/5^2=1#,

is clearly the standard form of equation of an ellipse with center at #(0,0)#, major axis as #5xx2=10# (at #y#-axis) and minor axis as #3xx2=6# (at #x#-axis). Clearly vertices are #(-5,0)# and #(5,0)#.

Eccentricity is #e=sqrt(1-3^2/5^2)=sqrt(1-9/25)=sqrt(16/25)=4/5=0.8#.

And hence focii are #(0,-be)# and #(0,be)# i.e. #(0,-4)# and #(0,4)#.

As it is an ellipse question of asymptote does not arise

The graph appears as below.

graph{x^2/9+y^2/25=1 [-10, 10, -5, 5]}