32. $$ \frac{x^{2}}{16}+\frac{y^{2}}{81}=1 $$

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UpStudy AI · Accepted Answer

The given equation is

$$
\frac{x^{2}}{16}+\frac{y^{2}}{81}=1
$$

This is the equation of an ellipse in standard form, where the general form of an ellipse is

$$
\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1
$$

From the given equation, we can identify:

- $$ a^{2} = 16 $$ which gives $$ a = \sqrt{16} = 4 $$
- $$ b^{2} = 81 $$ which gives $$ b = \sqrt{81} = 9 $$

### Characteristics of the Ellipse

1. **Center**: The center of the ellipse is at the origin $$(0, 0)$$.
2. **Semi-major axis**: The length of the semi-major axis is $$ b = 9 $$ (along the y-axis since $$ b > a $$).
3. **Semi-minor axis**: The length of the semi-minor axis is $$ a = 4 $$ (along the x-axis).
4. **Vertices**: The vertices along the y-axis are at $$(0, 9)$$ and $$(0, -9)$$.
5. **Co-vertices**: The co-vertices along the x-axis are at $$(4, 0)$$ and $$(-4, 0)$$.
6. **Foci**: The distance of the foci from the center is given by $$ c = \sqrt{b^{2} - a^{2}} = \sqrt{81 - 16} = \sqrt{65} $$. The foci are located at $$(0, c)$$ and $$(0, -c)$$, which are approximately $$(0, 8.06)$$ and $$(0, -8.06)$$.

### Summary

- Center: $$(0, 0)$$
- Semi-major axis: $$9$$ (along y-axis)
- Semi-minor axis: $$4$$ (along x-axis)
- Vertices: $$(0, 9)$$, $$(0, -9)$$
- Co-vertices: $$(4, 0)$$, $$(-4, 0)$$
- Foci: $$(0, \sqrt{65})$$, $$(0, -\sqrt{65})$$ or approximately $$(0, 8.06)$$, $$(0, -8.06)$$

If you need further analysis or specific calculations, please let me know!
The equation $$ \frac{x^{2}}{16} + \frac{y^{2}}{81} = 1 $$ represents an ellipse centered at the origin with a semi-major axis of 9 units along the y-axis and a semi-minor axis of 4 units along the x-axis. The vertices are at $$(0, 9)$$ and $$(0, -9)$$, and the co-vertices are at $$(4, 0)$$ and $$(-4, 0)$$. The foci are located at $$(0, \sqrt{65})$$ and $$(0, -\sqrt{65})$$, approximately at $$(0, 8.06)$$ and $$(0, -8.06)$$.

32. \( \frac{x^{2}}{16}+\frac{y^{2}}{81}=1 \)

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