How do you find the center, vertices, foci and eccentricity of $\frac{x^{2}}{36} + \frac{y^{2}}{64} = 1$ $x^2/36+y^2/64=1$ ?

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Answer 1 · 2016-07-22T10:01:14

(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 $e=sqrt7/4$ .

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

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 $A(a,0)=A(6,0), A'(-a,0)=A'(-6,0), B(0,b)=B(0,8), B'(0,-b)=B'(0,-8)$

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