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

Beginner2023-02-19Added 10 answers

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

Compared to the standard ellipse equation $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$,

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

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

The Eccentricity e is given by,

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

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

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

$S\prime (0,-be)=S\prime (0,-2\sqrt{7})$

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

The Vertices are $A(a,0)=A(6,0),A\prime (-a,0)=A\prime (-6,0),B(0,b)=B(0,8),B\prime (0,-b)=B\prime (0,-8)$

Compared to the standard ellipse equation $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$,

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

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

The Eccentricity e is given by,

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

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

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

$S\prime (0,-be)=S\prime (0,-2\sqrt{7})$

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

The Vertices are $A(a,0)=A(6,0),A\prime (-a,0)=A\prime (-6,0),B(0,b)=B(0,8),B\prime (0,-b)=B\prime (0,-8)$

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