How do you find the center, vertices, foci and eccentricity of #x^2/36+y^2/64=1#?

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Answer 1 · 2016-07-22T10:01:14

(1) Centre #C(0,0)#. graph{x^2/36+y^2/64=1 [-17.78, 17.78, -8.89, 8.89]}

(2) Vertices #A(6,0), A'(-6,0), B(0,8), B'(0,-8)#.

(3) Focii #S(0,2sqrt7), S'(0,-2sqrt7)#

(4) Eccentricity #e=sqrt7/4#.

#x^2/36+y^2/64=1#

Comparing with the std. eqn. of ellipse #x^2/a^2+y^2/b^2=1#,

#a^2=36, b^2=64 rArr b=8>6=a#

Therefore, the eqn. represents an Ellipse with Major axis along Y- axis & Minor Axis along X-axis

The Eccentricity e is given by,

#a^2=b^2(1-e^2)rArr36=64(1-e^2)rArr36/64=1-e^2#

#:. e^2=1-36/64=1-9/16=7/16rArr e=sqrt7/4#

The Focii are #S(0,be)=S(0,8*sqrt7/4)=S(0,2sqrt7)# &

#S'(0,-be)=S'(0,-2sqrt7)#

The Centre of the ellipse is #C(0,0)#.

The Vertices are #A(a,0)=A(6,0), A'(-a,0)=A'(-6,0), B(0,b)=B(0,8), B'(0,-b)=B'(0,-8)#