chaomeinguyzyp3

2023-02-18

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

Gerald Dickerson

$\frac{{x}^{2}}{36}+\frac{{y}^{2}}{64}=1$
Compared to the standard ellipse equation $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$,
${a}^{2}=36,{b}^{2}=64⇒b=8>6=a$
Therefore, the eqn. represents an Ellipse with Major axis along Y- axis & Minor Axis along X-axis
The Eccentricity e is given by,
${a}^{2}={b}^{2}\left(1-{e}^{2}\right)⇒36=64\left(1-{e}^{2}\right)⇒\frac{36}{64}=1-{e}^{2}$
$\therefore {e}^{2}=1-\frac{36}{64}=1-\frac{9}{16}=\frac{7}{16}⇒e=\frac{\sqrt{7}}{4}$
The Focii are $S\left(0,be\right)=S\left(0,8\cdot \frac{\sqrt{7}}{4}\right)=S\left(0,2\sqrt{7}\right)$ &
$S\prime \left(0,-be\right)=S\prime \left(0,-2\sqrt{7}\right)$
The Centre of the ellipse is $C\left(0,0\right)$.
The Vertices are $A\left(a,0\right)=A\left(6,0\right),A\prime \left(-a,0\right)=A\prime \left(-6,0\right),B\left(0,b\right)=B\left(0,8\right),B\prime \left(0,-b\right)=B\prime \left(0,-8\right)$

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