How do you find the foci, center, vertex for #4x^2+9y^2-16x+18y-11=0#?

Question

10510 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-13T16:32:56

Center is #(2,1)#,focii would be #(2+sqrt5, 1) and (2-sqrt5,1)#Vertices would be #(-1,1)# and #(5,1)#

#4(x^2 -4x) +9(y^2+2y)-11=0#

#4(x^2 -4x+4-4) +9(y^2 +2y +1-1) -11=0#

#4(x-2)^2 +9(y-1)^2 -16-9-11=0#

#(x-2)^2 +9(y-1)^2 =36#

#(x-2)^2 /3^2 +(y-1)^2 /2^2 =1# This represents an ellipse. Center is (2,1),

Distance from the centre to the focus is c= #sqrt(3^2-2^2)=sqrt5#

Hence focii would be #(2+sqrt5, 1) and (2-sqrt5,1)#

Vertices would be #(-1,1)# and #(5,1)#