How do you find the center, vertices, foci and eccentricity of x^2/36+y^2/64=1x236+y264=1?

1 Answer
Jul 22, 2016

(1) Centre C(0,0)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.

Explanation:

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)