Pre-Calculus Ellipses Worksheet with detailed solutions and graphs.

It looks like you've uploaded a Pre-Calculus worksheet on ellipses, and it appears to be a completed key (answer sheet) for graphing ellipses, identifying their centers, vertices, co-vertices (CV), and foci.

Let me walk through the solution process for each problem, explaining how to solve them step by step. This will help reinforce understanding even though the answers are already filled in.

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🔷 General Form of an Ellipse

The standard form of an ellipse is:

- Horizontal major axis:
$$
\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \quad \text{where } a > b
$$
- Vertical major axis:
$$
\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \quad \text{where } a > b
$$

Where:
- $(h,k)$ is the center
- $a$ = semi-major axis length
- $b$ = semi-minor axis length
- $c = \sqrt{a^2 - b^2}$ → distance from center to each focus
- Vertices: along major axis, at distance $a$ from center
- Co-vertices (CV): along minor axis, at distance $b$ from center
- Foci: along major axis, at distance $c$ from center

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## ✔ Problem-by-Problem Explanation

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1. $\frac{x^2}{16} + \frac{y^2}{4} = 1$

- This is centered at $(0,0)$ since no $h,k$ shifts.
- Compare: $\frac{x^2}{16} + \frac{y^2}{4} = 1$ → $a^2 = 16$, $b^2 = 4$
- So $a = 4$, $b = 2$
- Since $a^2$ is under $x^2$, the major axis is horizontal.
- $c^2 = a^2 - b^2 = 16 - 4 = 12$ → $c = \sqrt{12} = 2\sqrt{3}$

✔ Center: $(0,0)$
✔ Vertices (along x-axis): $(\pm a, 0) = (\pm 4, 0)$
✔ Co-vertices (along y-axis): $(0, \pm b) = (0, \pm 2)$
✔ Foci: $(\pm c, 0) = (\pm 2\sqrt{3}, 0)$

✔️ Graph: Horizontal ellipse centered at origin.

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2. $\frac{x^2}{9} + \frac{y^2}{36} = 1$

- Center: $(0,0)$
- $a^2 = 36$, $b^2 = 9$ → $a = 6$, $b = 3$
- Since $a^2$ is under $y^2$, major axis is vertical
- $c^2 = a^2 - b^2 = 36 - 9 = 27$ → $c = \sqrt{27} = 3\sqrt{3}$

✔ Center: $(0,0)$
✔ Vertices (up/down): $(0, \pm 6)$
✔ Co-vertices (left/right): $(\pm 3, 0)$
✔ Foci: $(0, \pm 3\sqrt{3})$

✔️ Graph: Vertical ellipse centered at origin.

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3. $25x^2 + 4y^2 = 100$

- Divide both sides by 100:
$$
\frac{25x^2}{100} + \frac{4y^2}{100} = 1 \Rightarrow \frac{x^2}{4} + \frac{y^2}{25} = 1
$$
- Now: $a^2 = 25$, $b^2 = 4$ → $a = 5$, $b = 2$
- Major axis is vertical (since $y^2$ term has larger denominator)
- $c^2 = a^2 - b^2 = 25 - 4 = 21$ → $c = \sqrt{21}$

✔ Center: $(0,0)$
✔ Vertices: $(0, \pm 5)$
✔ Co-vertices: $(\pm 2, 0)$
✔ Foci: $(0, \pm \sqrt{21})$

✔️ Graph: Vertical ellipse centered at origin.

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4. $7x^2 = 35 - 5y^2$

- Rearranging: $7x^2 + 5y^2 = 35$
- Divide both sides by 35:
$$
\frac{7x^2}{35} + \frac{5y^2}{35} = 1 \Rightarrow \frac{x^2}{5} + \frac{y^2}{7} = 1
$$
- So: $a^2 = 7$, $b^2 = 5$ → $a = \sqrt{7}$, $b = \sqrt{5}$
- Major axis is vertical (since $y^2$ has larger denominator)
- $c^2 = a^2 - b^2 = 7 - 5 = 2$ → $c = \sqrt{2}$

✔ Center: $(0,0)$
✔ Vertices: $(0, \pm \sqrt{7})$
✔ Co-vertices: $(\pm \sqrt{5}, 0)$
✔ Foci: $(0, \pm \sqrt{2})$

✔️ Graph: Vertical ellipse, smaller than previous ones.

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5. $\frac{(x-2)^2}{9} + \frac{(y-1)^2}{4} = 1$

- This is shifted: center at $(h,k) = (2,1)$
- $a^2 = 9$, $b^2 = 4$ → $a = 3$, $b = 2$
- Since $a^2$ is under $x^2$, horizontal major axis
- $c^2 = a^2 - b^2 = 9 - 4 = 5$ → $c = \sqrt{5}$

✔ Center: $(2,1)$
✔ Vertices: Move $a=3$ left/right from center:
- $(2 \pm 3, 1) = (5,1), (-1,1)$
✔ Co-vertices: Move $b=2$ up/down:
- $(2, 1 \pm 2) = (2,3), (2,-1)$
✔ Foci: Move $c=\sqrt{5}$ left/right:
- $(2 \pm \sqrt{5}, 1)$

✔️ Graph: Horizontal ellipse centered at $(2,1)$

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6. $\frac{(x-4)^2}{9} + \frac{(y+2)^2}{25} = 1$

- Center: $(h,k) = (4, -2)$
- $a^2 = 25$, $b^2 = 9$ → $a = 5$, $b = 3$
- Since $a^2$ is under $y^2$, vertical major axis
- $c^2 = a^2 - b^2 = 25 - 9 = 16$ → $c = 4$

✔ Center: $(4, -2)$
✔ Vertices: Move $a=5$ up/down:
- $(4, -2 \pm 5) = (4,3), (4,-7)$
✔ Co-vertices: Move $b=3$ left/right:
- $(4 \pm 3, -2) = (7,-2), (1,-2)$
✔ Foci: Move $c=4$ up/down:
- $(4, -2 \pm 4) = (4,2), (4,-6)$

✔️ Graph: Vertical ellipse centered at $(4,-2)$

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📌 Summary Table

| Problem | Equation | Center | Orientation | Vertices | Co-vertices | Foci |
|--------|---------|--------|-------------|----------|--------------|-------|
| 1 | $\frac{x^2}{16} + \frac{y^2}{4} = 1$ | (0,0) | Horizontal | (±4,0) | (0,±2) | (±2√3, 0) |
| 2 | $\frac{x^2}{9} + \frac{y^2}{36} = 1$ | (0,0) | Vertical | (0,±6) | (±3,0) | (0,±3√3) |
| 3 | $\frac{x^2}{4} + \frac{y^2}{25} = 1$ | (0,0) | Vertical | (0,±5) | (±2,0) | (0,±√21) |
| 4 | $\frac{x^2}{5} + \frac{y^2}{7} = 1$ | (0,0) | Vertical | (0,±√7) | (±√5,0) | (0,±√2) |
| 5 | $\frac{(x-2)^2}{9} + \frac{(y-1)^2}{4} = 1$ | (2,1) | Horizontal | (5,1), (-1,1) | (2,3), (2,-1) | (2±√5, 1) |
| 6 | $\frac{(x-4)^2}{9} + \frac{(y+2)^2}{25} = 1$ | (4,-2) | Vertical | (4,3), (4,-7) | (7,-2), (1,-2) | (4,2), (4,-6) |

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✔ Final Notes:

- Always rewrite equations in standard form.
- Identify whether the major axis is horizontal or vertical based on which denominator is larger.
- Use $c^2 = a^2 - b^2$ to find foci.
- Shift the center accordingly if there are $h$ and $k$ terms.
- Plot the center first, then add vertices, co-vertices, and foci.

If you'd like, I can generate a clean version of this worksheet with all steps clearly labeled, or help you practice similar problems!