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

Jul 22, 2016

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

(2) Vertices $A \left(6 , 0\right) , A ' \left(- 6 , 0\right) , B \left(0 , 8\right) , B ' \left(0 , - 8\right)$.

(3) Focii $S \left(0 , 2 \sqrt{7}\right) , S ' \left(0 , - 2 \sqrt{7}\right)$

(4) Eccentricity $e = \frac{\sqrt{7}}{4}$.

#### Explanation:

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

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

${a}^{2} = 36 , {b}^{2} = 64 \Rightarrow b = 8 > 6 = a$

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

The Eccentricity e is given by,

${a}^{2} = {b}^{2} \left(1 - {e}^{2}\right) \Rightarrow 36 = 64 \left(1 - {e}^{2}\right) \Rightarrow \frac{36}{64} = 1 - {e}^{2}$

$\therefore {e}^{2} = 1 - \frac{36}{64} = 1 - \frac{9}{16} = \frac{7}{16} \Rightarrow e = \frac{\sqrt{7}}{4}$

The Focii are $S \left(0 , b e\right) = S \left(0 , 8 \cdot \frac{\sqrt{7}}{4}\right) = S \left(0 , 2 \sqrt{7}\right)$ &

$S ' \left(0 , - b e\right) = S ' \left(0 , - 2 \sqrt{7}\right)$

The Centre of the ellipse is $C \left(0 , 0\right)$.

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)