Ellipse and its properties

class: center, middle, inverse, title-slide

# Ellipse and its properties
## Analytic Geometry
### Academy of Mathematics
### High School UPAEP Angelópolis
### Fall 2022

---

class: inverse, center, middle

# Geometrical elements of an ellipse

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class: middle

# Geometrical elements of an ellipse (1/9)

---
# Geometrical elements of an ellipse (2/9)

## Warning!!
 The position of the geometrical elements **depend on the position of the ellipse**: horizontal or vertical.

.pull-left[

<img src="horizontal-ellipse.png" width="110%" style="display: block; margin: auto;" />
]

.pull-right[

<img src="vertical-ellipse.png" width="110%" style="display: block; margin: auto;" />
]
---
# Geometrical elements of an ellipse (3/9)

.pull-left[

The elements of an ellipse are:

- The **center** `$\color{red}{C}(\color{green}{h},\color{purple}{k})$`: it is a point in the plane

- The **major axis**: it is the largest segment that passes through the center.

- The **vertices** `$\color{orange}{V_1}$` and `$\color{orange}{V_2}$`: they are the _endpoints_ of the major axis.
 
]

.pull-right[

<img src="horizontal-ellipse.png" width="130%" style="display: block; margin: auto;" />
]

---
# Geometrical elements of an ellipse (4/9)

.pull-left[
 We define
$$
`\begin{aligned}
{\large \color{orange}{a}} & = \text{distance between the}\\
 & \, \,\,\quad \text{center and a vertex}\\
 & = d(\color{red}{C}, \color{orange}{V_1}) \\
 & = d(\color{red}{C}, \color{orange}{V_2})
\end{aligned}`
$$
- Taking in consideration the previous value we get that **the length of the major axis** is `${\large \color{darkorange}{2a}}$`.
]

.pull-right[

<img src="horizontal-ellipse.png" width="130%" style="display: block; margin: auto;" />
]

---
# Geometrical elements of an ellipse (5/9)

.pull-left[

- The **minor axis**: it is the shortest segment that passes through the center.

- The **covertices** `$\color{darkcyan}{B_1}$` and `$\color{darkcyan}{B_2}$`: they are the _endpoints_ of the minor axis.

]

.pull-right[

<img src="horizontal-ellipse.png" width="130%" style="display: block; margin: auto;" />
]

---
# Geometrical elements of an ellipse (6/9)

.pull-left[

We define
$$
`\begin{aligned}
{\large \color{darkcyan}{b}} & = \text{distance between the}\\
 & \, \,\,\quad \text{center and a covertex}\\
 & = d(\color{red}{C}, \color{darkcyan}{B_1}) \\
 & = d(\color{red}{C}, \color{darkcyan}{B_2})
\end{aligned}`
$$

- Taking in consideration the previous value we get that **the length of the minor axis** is `${\large \color{darkcyan}{2b}}$`.
]

.pull-right[

<img src="horizontal-ellipse.png" width="130%" style="display: block; margin: auto;" />
]

---
# Geometrical elements of an ellipse (7/9)

.pull-left[

The **foci** `$\color{blue}{F_1}$` and `$\color{blue}{F_2}$` are two points ON the major axis.

To find the location of the foci, we consider the following distance:
$$
`\begin{aligned}
{\large \color{blue}{c}} & = \text{distance between the}\\
 & \, \,\,\quad \text{center and a focus}\\
 & = d(\color{red}{C}, \color{blue}{F_1}) \\
 & = d(\color{red}{C}, \color{blue}{F_2})
\end{aligned}`
$$
]

.pull-right[

<img src="horizontal-ellipse.png" width="130%" style="display: block; margin: auto;" />
]

---
# Geometrical elements of an ellipse (8/9)

.pull-left[

How `$\color{darkorange}{a}$`, `$\color{darkcyan}{b}$`, and `$\color{blue}{c}$` are related?

$$
 \color{blue}{c}^{2} = \color{darkorange}{a}^{2} - \color{darkcyan}{b}^{2}
$$
]

.pull-right[

<img src="horizontal-ellipse.png" width="130%" style="display: block; margin: auto;" />
]

---
class: middle

# Geometrical elements of an ellipse (9/9)

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class: inverse, center

PART I

Standard equation of the ellipse

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# Ellipse in standard form (1/2)

What do we need to know to find the standard equation of an ellipse?

.pull-left[

- The **center** `$\color{red}{C}(\color{green}{h}, \color{purple}{k})$`.

- The value `${\large \color{darkorange}{a}}$` (called semi-major axis).

]

.pull-right[

- The value `${\large\color{darkcyan}{b}}$` (called semi-minor axis).

- The **position** of the ellipse (horizontal or vertical)

]

---

# Ellipse in standard form (2/2)

## Horizontal ellipse

$$
\frac{(x-\color{green}{h})^{2}}{\color{darkorange}{a}^{2}} + \frac{(y-\color{purple}{k})^{2}}{\color{darkcyan}{b}^{2}} = 1
$$

<hr>

## Vertical ellipse

$$
\frac{(x-\color{green}{h})^{2}}{\color{darkcyan}{b}^{2}} + \frac{(y-\color{purple}{k})^{2}}{\color{darkorange}{a}^{2}} = 1
$$

---

# Example 1 (1/2)

**Problem 1**

Find the standard equation of the horizontal ellipse centered at `$\color{red}{C}(\color{green}{5}, \color{purple}{-7})$`, with major axis of length `$18$`, and minor axis of length `$6$`.

**Solution**

First of all, let's write the information given in the statement of the problem:

$$
`\begin{aligned}
\text{center: }& C(\color{green}{5},\color{purple}{-7}), \\
\text{major axis: }& \color{darkorange}{2a}=\color{darkorange}{18} \\
\text{minor axis: }& \color{darkcyan}{2b}=\color{darkcyan}{6} \\
\text{position: }& \textbf{horizontal}
\end{aligned}`
$$

---
# Example 1(2/2)

Then, we get `$\color{darkorange}{a} = \frac{18}{2} = \color{darkorange}{9}$`, and `$\color{darkcyan}{b}= \frac{6}{2} = \color{darkcyan}{3}$`.

So, after subtituting into the standard equation (for the horizontal case), and reducing terms, we get: $$
`\begin{gathered}
\frac{(x - \color{green}{5} )^{2}}{(\color{darkorange}{9})^{2}} + \frac{(y-(\color{purple}{-7}))^{2}}{(\color{darkcyan}{3})^{2}} = 1 \\
\mbox{} \\
\frac{(x - \color{green}{5} )^{2}}{\color{darkorange}{81}} + \frac{(y+\color{purple}{7})^{2}}{\color{darkcyan}{9}} = 1 
\end{gathered}`
$$

---

# Example 2 (1/2)

**Problem 2**

Find the standard equation of the vertical ellipse centered at `$\color{red}{C}(\color{green}{0}, \color{purple}{4})$`, with major axis of length `$10$`, and minor axis of length `$2$`.

**Solution**

Again, let's write the information given in the statement of the problem:

$$
`\begin{aligned}
\text{center: }& C(\color{green}{0},\color{purple}{4}), \\
\text{major axis: }& \color{darkorange}{2a}=\color{darkorange}{10} \\
\text{minor axis: }& \color{darkcyan}{2b}=\color{darkcyan}{2} \\
\text{position: }& \textbf{vertical}
\end{aligned}`
$$

---
# Example 2 (2/2)

Then, we get `$\color{darkorange}{a} = \frac{10}{2} = \color{darkorange}{5}$`, and `$\color{darkcyan}{b}= \frac{2}{2} = \color{darkcyan}{1}$`.

So, after subtituting into the standard equation (for the vertical case), and reducing terms, we get: $$
`\begin{gathered}
\frac{(x - \color{green}{0} )^{2}}{(\color{darkcyan}{1})^{2}} + \frac{(y-\color{purple}{4})^{2}}{(\color{darkorange}{5})^{2}} = 1 \\
\mbox{} \\
\frac{x^{2}}{\color{darkcyan}{1}} + \frac{(y-\color{purple}{4})^{2}}{\color{darkorange}{25}} = 1 \\
\mbox{} \\
x^{2} + \frac{(y-\color{purple}{4})^{2}}{\color{darkorange}{25}} = 1
\end{gathered}`
$$

---

# Example 3

.pull-left[
 **Problem 3**

Find the standard equation of the ellipse in the figure.

<img src="ellipse02.png" width="100%" style="display: block; margin: auto;" />
]

.pull-right[
 **Solution**

We note that `$\color{red}{C}(\color{green}{4}, \color{purple}{-3})$`. In addition, the position of the ellipse is vertical, and we identify that `$\color{darkorange}{a} = \color{darkorange}{3}$`, and `$\color{darkcyan}{b}=\color{darkcyan}{2}$`. So, we have the 4 elements to calculate the standard equation, and after the subtitution we obtain $$
`\begin{aligned}
\frac{(x-\color{green}{4})^{2}}{(\color{darkcyan}{2})^{2}} + \frac{(y-(\color{purple}{-3}))^{2}}{(\color{darkorange}{3})^{2}} = 1 \\
\mbox{} \\
\frac{(x-\color{green}{4})^{2}}{\color{darkcyan}{4}} + \frac{(y+\color{purple}{3})^{2}}{\color{darkorange}{9}} = 1
\end{aligned}`
$$

]

---
class: inverse, center

PART II

General equation of the circle

---
# From the standard equation to the general equation

Given an ellipse, to determine its **general equation**, we have to follow the next steps:

- **Step 1**: Find the standard equation of the circle.

- **Step 2**: Expand the binomials that appear.

- **Step 3**: Add the fractions. Reduce terms.
 
 - **Step 4**: "Pass" the denominator to the right hand.
 
 - **Step 5**: "Pass" the constant on the right hand to the left side and simplify.

---
# Example 4(1/3)

.pull-left[
 **Problem 4**

Find the general equation of the ellipse in the figure.

]

.pull-right[
 **Solution**

**Step 1**: In the **Example 3** we saw that the standard equation of the ellipse is:

$$
\frac{(x-\color{green}{4})^{2}}{\color{darkcyan}{4}} + \frac{(y+\color{purple}{3})^{2}}{\color{darkorange}{9}} = 1
$$

**Step 2**: Now, we expand the binomials to get

$$
\frac{x^2 - 8x + 16}{4} + \frac{y^2 + 6x + 9}{9} = 1
$$
]

---
# Example 4 (2/3)

**Step 3**: Now, let's add the fractions. We procced as follows:

$$
`\begin{gathered}
\frac{x^2 - 8x + 16}{4} + \frac{y^2 + 6x + 9}{9} = 1 \\
\\
\frac{9(x^2 - 8x +16) + 4(y^2 + 6x + 9)}{36} = 1 \\
\\
\frac{9x^2 -72x + 144 + 4y^2 +24y + 36}{36} = 1 \\
\\
\frac{9x^2 + 4y^2 -72x + 24x + 180}{36} = 1
\end{gathered}`
$$
---
# Example 4 (3/3)

**Step 4**: Next, we "pass" the denominator to the left hand, this is: $$
`\begin{gathered}
\frac{9x^2 + 4y^2 -72x + 24x + 180}{36} = 1 \\
\\
9x^2 + 4y^2 -72x + 24x + 180 = 36
\end{gathered}`
$$

**Step 5**: Finally, we reduce the constants: $$
`\begin{gathered}
9x^2 + 4y^2 -72x + 24x + 180 = 36 \\
\\
9x^2 + 4y^2 -72x + 24x + 180 - 36 = 0 \\
\\
\color{red}{9x^2 + 4y^2 -72x + 24x + 144 = 0}
\end{gathered}`
$$

---
class: inverse, center

PART III

Getting the key features of the ellipse from an equation

---
# From the standard equation

.pull-left[
 Let's consider the equation $$
\frac{(x+4)^2}{4} + \frac{(y-5)^2}{9} = 1
$$

It looks like the **standard equation of an ellipse**: then, at least, we _look for_ the 4 key elements of an ellipse:

- the **center**
 - the **position** (horizontal or vertical)
 - the **semi-major axis** `${\large \color{darkorange}{a}}$`
 - the **semi-minor axis** `${\large \color{darkcyan}{b}}$`
]

.pull-right[
 To find the elements we have to:

- For the **center**: take the numbers with opposite sign that appears next to the variables `$x$` and `$y$`, respectively.
 - For the **position**: it is indicated by the variable over the largest denominator.
 - For the value of `${\large \color{darkorange}{a}}$`: take the square root of the largest denominator.
 - For the value of `${\large \color{darkcyan}{b}}$`: take the square root of the smallest denominator.

]

---
# Example 5 (1/2)

**Problem 5**

Find the elements of the figure represented by the following equation $$
\frac{(x+4)^2}{4} + \frac{(y-5)^2}{9} = 1
$$

**Solution**

We note that it is the standard equation of an ellipse.

- The **center** of the ellipse is `$\color{red}{C}(\color{green}{-4}, \color{purple}{5})$`. Here, we just took the numbers next to `$x$` and `$y$`, respectively, with the opposite sign.

- Because the largest denominator is `$\color{darkorange}{9}$`, and it is under the variable `$\textbf{y}$`, the ellipse is in **vertical position**.

---

# Example 5 (2/2)

- The largest denominator is `$\color{darkorange}{9}$`, then, we have that `$\color{darkorange}{a} = \sqrt{9} = \color{darkorange}{3}$`.

- The smallest denominator is `$\color{darkcyan}{4}$`, then, we have that `$\color{darkcyan}{b} = \sqrt{4} = \color{darkcyan}{2}$`.

Then, the final answer is $$
`\begin{aligned}
\text{center: }& C(\color{green}{-4},\color{purple}{5}), \\
\text{position: }& \textbf{vertical} \\
\text{semi-major axis: }& \color{darkorange}{a}=\color{darkorange}{3} \\
\text{semi-minor axis: }& \color{darkcyan}{b}=\color{darkcyan}{2}
\end{aligned}`
$$
---
# Example 6

**Problem 6**

Find the elements of the figure represented by the following equation $$
\frac{ ( x-2 )^2}{25} + \frac{ ( y+5 )^2 }{5} = 1
$$

**Solution**

We note that it is a standard equation of an ellipse. In this case, the final answer is $$
`\begin{aligned}
\text{center: }& C(\color{green}{2},\color{purple}{-5}), \\
\text{position: }& \textbf{horizontal} \\
\text{semi-major axis: }& \color{darkorange}{a}=\color{darkorange}{5} \\
\text{semi-minor axis: }& \color{darkcyan}{b}=\color{darkcyan}{\sqrt{5}}
\end{aligned}`
$$

---
# From the general equation to the standard equation

Consider the equation given by

$$\color{darkred}{A}x^2 + \color{darkblue}{B}y^2+Cx+Dy+F=0
$$

If `$\color{darkred}{A}$` and `$\color{darkblue}{B}$` **have the same sign**, then we _expect_ a **general equation of an ellipse**.

In this case, the strategy is **transforming into the standard equation** by _completing squares_ for both `$x$` and `$y$`: later we just apply the strategy in the case of the standard equation.

---
# Example 7 (1/4)

**Problem 7**

Find the key elements of the figure with equation given by

$$
4x^2 + 25y^2 - 32x + 200y + 364 = 0
$$

**Solution**

We note that the coefficients of `$x^2$` and `$y^2$` have the same sign, then we _expect_ that it is the **general equation of an ellipse**. Then, the key elements that we are looking for are the **center**, the **position**, the **semi-major axis** `$\color{darkorange}{a}$`, and the **semi-minor axis** `$\color{darkcyan}{b}$`.

---
# Example 7 (2/4)

**Step 1**: Group the terms with the same variable, and move the constant to the right hand.

$$
(\color{brown}{4x^2 - 32x}) + (\color{blue}{25y^2 + 200y }) = \textbf{-364}
$$

**Step 2**: Factor out the coefficients of squared terms.

$$
\color{brown}{4}(x^2 - 8x) + \color{blue}{25}(y^2 + 8y ) = -364
$$

**Step 3**: Make space for values on both sides to complete the squares.

$$
\color{brown}{4}(x^2 - 8x \color{red}{+}\,\,\,\,\, ) + \color{blue}{25}(y^2 + 8y \color{red}{+} \,\,\,\,\, ) = -364 \color{red}{+} \color{brown}{4} (\,\,\,\,\, ) \color{red}{+} \color{blue}{25}(\,\,\,\,\, )
$$

---
# Example 7 (3/4)

**Step 4**: Complete the perfect trinomial squares. Add the same values to the right side.

$$
4(x^2 - 8x + \color{brown}{16} ) + 25(y^2 + 8y + \color{blue}{16} ) = -364 + 4( \color{brown}{16} ) + 25( \color{blue}{16} )
$$

**Step 5**: Multiply and add.

$$
4(x^2 - 8x + 16 ) + 25(y^2 + 8y + 16 ) = \textbf{100}
$$

**Step 6**: Rewrite trinomials as squared binomials.

$$
4(x - 4)^{2} + 25(y + 4)^{2} = 100
$$
---
# Example 7 (4/4)

**Step 7**: Divide through by the constant on the right side to make it `$1$`.

$$
\frac{ 4(x - 4)^{2} }{100} + \frac{ 25(y + 4)^{2} }{100} = \frac{100}{100}
$$

**Step 8**: Simplify fractions.

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

**Final answer**: the key elements are $$
`\begin{aligned}
\text{center: }& C(\color{green}{4},\color{purple}{-4}), \\
\text{position: }& \textbf{horizontal} \\
\text{semi-major axis: }& \color{darkorange}{a}=\color{darkorange}{5} \\
\text{semi-minor axis: }& \color{darkcyan}{b}=\color{darkcyan}{2}
\end{aligned}`
$$

---
class: inverse, center

PART IV

Applications

---
# Example 8 (1/8)

Find all the geometrical elements of the ellipse with foci `$(15,-2)$` and `$(3,-2)$`, and covertices `$(9,6)$` and `$(9,-10)$`. Sketch this ellipse on the Cartesian plane.

**Solution**

As we have seen, all the elements depend on `$4$` of the geometrical elements of the ellipse:

- the **center**
 - the value of `$\color{darkorange}{a}$`
 - the value of `$\color{darkcyan}{b}$`
 - the **position** of the ellipse

So, let's find all of them.

---
# Example 8 (2/8)

## The center

We know that the center is the **midpoint of the minor axis**, but it is also the **midpoint of the segment that joins the foci**.

.pull-left[
 The foci are `$(15,-2)$` and `$(3, -2)$`, then, the midpoint is $$
`\begin{gathered}
C\left( \frac{15 + 3}{2}, \frac{-2 + (-2)}{2} \right) \\
C\left( \frac{18}{2}, \frac{-4}{2} \right) \\
C(\color{green}{9}, \color{purple}{-2})
\end{gathered}`
$$

]

.pull-right[
 The _covertices_ `$(9,6)$` and `$(9,-10)$` are the _endpoints_ of the _minor axis_, then, the midpoint is found as $$
`\begin{gathered}
C\left( \frac{9 + 9}{2}, \frac{6 + (-10)}{2} \right) \\
C\left( \frac{18}{2}, \frac{-4}{2} \right) \\
C(\color{green}{9}, \color{purple}{-2})
\end{gathered}`
$$

]

.center[
 In both cases we got that the **center** is `$C(\color{green}{9}, \color{purple}{-2})$`.
]

---

# Example 8 (3/8)

## The value of `$\color{darkcyan}{b}$`

In this problem we are given with the covertices `$B_1(9,6)$` and `$B_2(9,-10)$`, this is, the endopoints of the minor axis, then, we can get the **length of the minor axis**, which is equal to `$\color{darkcyan}{2b}$`. We have that $$
 \color{darkcyan}{2b} = d(B_1 , B_2 ) = 16,
$$
and from here, $$
 \color{darkcyan}{b} = \frac{16}{2} = 8.
$$

---
# Example 8 (4/8)

## The value of `$\color{blue}{c}$`

In the statement of this problem we are given with both **foci** `$F_1(15, -2)$` and `$F_2(3,-2)$`. Using this information there are two possibilities to find the value of `$\color{blue}{c}$`.

.pull-left[
 Because the distance from the center to a focus is `$\color{blue}{c}$`, we know that the **distance between both foci** is `$\color{blue}{2c}$`. In this case: $$
 \color{blue}{2c} = d (F_1, F_2) = 12
$$
]

.pull-right[
 We know that the center is `$C(9,-2)$`, then $$
`\begin{gathered}
 \color{blue}{c} = d(F_1, C) = 6 \\
 \color{blue}{c} = d(F_2, C) = 6
\end{gathered}`
$$

]

.center[
 In both cases we get that `${\large \color{blue}{c} = \color{blue}{6}}$`.
]

---
# Example 8 (5/8)

## The value of `$\color{darkorange}{a}$`

We already know that `$\color{darkcyan}{b} = \color{darkcyan}{8}$` and `$\color{blue}{c} = \color{blue}{6}$`. Furthermore, we know that $$
 \color{blue}{c}^{2} = \color{darkorange}{a}^2 - \color{darkcyan}{b}^2,
$$
then, isolating `$a$` we get $$
 \color{darkorange}{a}^2 = \color{darkcyan}{b}^2 + \color{blue}{c}^{2}
$$

Making subtitution we get $$
 \color{darkorange}{a}^2 = \color{darkcyan}{8}^2 + \color{blue}{6}^2 = 64 + 36 = 100
$$
this is $$
{\large \color{darkorange}{a} = \color{darkorange}{10}}.
$$

---
# Example 8 (6/8)

## The position of the ellipse

In this problem we have at least two possibilities to look for position of the ellipse.

.pull-left[
 We note that the coordinates of both **foci** `$F_1(15, -2)$` and `$F_2(3,-2)$` only change in the first coordinate, this is, the position of the major axis (or the ellipse) has to be horizontal.
]

.pull-right[
 We know that the coordinates of both covertices `$(9,6)$` and `$(9,-10)$` are the endpoints of the minor axis, and they only change in the second coordinate. So the position of the minor axis is vertical, this is, the position of the major axis is horizontal.. 
]

.center[
 In both cases, the position of the ellipse is **horizontal**.
]

---
# Example 8 (7/8)

The missing elements are the length of the major and minor axes, the vertices, and the standard equation. We get that $$
 \color{darkorange}{2a} = \color{darkorange}{20} \text{ and } \color{darkcyan}{2b} = \color{darkcyan}{16}.
$$
On the other hand, the vertices are $$
 \color{darkorange}{V_1}( 19,-2) \text{ and } \color{darkorange}{V_2} = (-1, -2).
$$
The standard equation is $$
\frac{(x-9)^2}{100} + \frac{(y+2)^2}{64} = 1. 
$$
Finally, we sketch the obtained ellipse on the next figure.

---
class: clear

---
# Example 9 (1/4)

Find all the geometrical elements of the figure represented by $$
\frac{(x+2)^2}{4} + \frac{(y+3)^2}{9} = 1.
$$
Sketch this figure on the Cartesian plane.

**Solution**

We note that it is the standard equation of an ellipse.

- The **center** of the ellipse is `$\color{red}{C}(\color{green}{-2}, \color{purple}{-3})$`. Here we just took the numbers next to `$x$` and `$y$`, respectively, with the opposite sign.
 
 - Because the largest denominator is `$\color{darkorange}{9}$`, and it is under the variable `$\textbf{y}$`, the ellipse is in **vertical position**.

---
# Example 9 (2/4)

- The largest denominator is `$\color{darkorange}{9}$`, then, we have that `$\color{darkorange}{a} = \sqrt{9} = \color{darkorange}{3}$`. This implies that the **length of the major axis** is `$\color{darkorange}{2a} = \color{darkorange}{6}$`. 
 
 - The smallest denominator is `$\color{darkcyan}{4}$`, then, we have that `$\color{darkcyan}{b} = \sqrt{4} = \color{darkcyan}{2}$`. So, we get that the **length of the minor axis** is `$\color{darkcyan}{2b} = \color{darkcyan}{4}$`.
 
 - From the previous information, and using the fact that `$\color{blue}{c}^{2} = \color{darkorange}{a}^2 - \color{darkcyan}{b}^2$`, we obtain that $$
 \color{blue}{c}^2 = \color{darkorange}{9} - \color{darkcyan}{4} = 5
 $$
this is $$
 \color{blue}{c} = \color{blue}{\sqrt{5}}
$$

---
class: clear

---
# Example 9 (4/4)

We know that the missing elements are both vertices, both covertices, and both foci. As we can identify from the previous figure, their coordinates are the following ones:

- **vertices**: `$\color{darkorange}{V_1(-2, 0)}$` and `$\color{darkorange}{V_2(-2, -6)}$`. Here we use that `$\color{darkorange}{a}$` is the distance between the center `$\color{red}{C}(\color{green}{-2}, \color{purple}{-3})$` and a vertex.
 
 - **covertices**: `$\color{darkcyan}{B_1(0, -3)}$` and `$\color{darkcyan}{B_2(-4, -3)}$`. Here we use that `$\color{darkcyan}{b}$` is the distance between the center `$\color{red}{C}(\color{green}{-2}, \color{purple}{-3})$` and a covertex.

- **foci**: `$\color{blue}{F_1(-2, -3 + \sqrt{5}) }$` and `$\color{blue}{F_2 (-2, -3 -\sqrt{5}) }$`. Here we use that `$\color{blue}{c}$` is the distance between the center `$\color{red}{C}(\color{green}{-2}, \color{purple}{-3})$` and a focus.

---
# Example 10 (1/6)

Find all the geometrical elements of the figure represented by $$
9x^2 + 16y^2 -18x - 64 y -71 = 0.
$$
Sketch this figure on the Cartesian plane.

**Solution**

We note that the coefficients of `$x^2$` and `$y^2$` have the same sign, then we _expect_ that it is the **general equation of an ellipse**.

Now, let's try to transform the general equation into the standard equation of an ellipse to get all the elements.

---
# Example 10 (2/6)

First we group the terms with the same variable $$
 (\color{brown}{9x^2 - 18x}) + (\color{blue}{16y^2 - 64y}) = \textbf{71}
$$
then we factorize the coefficients of `$x^2$` and `$y^2$`, respectively, $$
 \color{brown}{9} (x^2 - 2x) + \color{blue}{16} (y^2 - 4y) = 71
$$
next we complete the perfect square trinomials on the left side, and we add the same values to the right side $$
`\begin{aligned}
 9 (x^2 - 2x + \color{brown}{1}) + 16 (y^2 - 4y + \color{blue}{4}) & = 71 + 9 (\color{brown}{1}) + 16 (\color{blue}{4}) \\
 9 (x^2 - 2x + 1) + 16 (y^2 - 4y + 4) & = 71 + 9 + 64 \\
 9 (x^2 - 2x + 1) + 16 (y^2 - 4y + 4) & = 144
\end{aligned}`
$$

---
# Example 10 (3/6)

Now, we rewrite the trinomials as squared binomials on the left side, and multiply and add on the right side $$
 9 (x - 1)^2 + 16 (y - 2)^2 = 144
$$
Next, we divide through by the constant on the right side to make it `$1$`: $$
 \frac{ 9(x-1)^2 }{144} + \frac{ 16 (y-2)^2 }{144} = \frac{144}{144}
$$
and we simplify each fraction to get the standard equation: $$
 \frac{ (x-1)^2 }{16} + \frac{ (y-2)^2 }{9} = 1
$$

---
# Example 10 (4/6)

Let's use the standard equation to get all the elements of the ellipse:

- The **center** is `$\color{red}{C}(\color{green}{1}, \color{purple}{2})$`.
 - Because the largest denominator is `$\color{darkorange}{16}$`, and it is under the variable `$\textbf{x}$`, the position of the ellipse is **horizontal**.
 - The value of `$\color{darkorange}{a}^2 = \color{darkorange}{16}$` implies that `$\color{darkorange}{a}=\color{darkorange}{4}$`. In particular, the **length of the major axis** is `$\color{darkorange}{2a} = \color{darkorange}{8}$`.
 - The value of `$\color{darkcyan}{b}^2 = \color{darkcyan}{9}$` implies that `$\color{darkcyan}{a}=\color{darkcyan}{3}$`. In particular, the **length of the major axis** is `$\color{darkcyan}{2b} = \color{darkcyan}{6}$`.
 - Now, we can get the value of `$\color{blue}{c}$` using the equation `$\color{blue}{c}^2 = \color{darkorange}{a}^2 - \color{darkcyan}{b}^2$`. In this case we obtain that `$\color{blue}{c}^2 = \color{darkorange}{16} - \color{darkcyan}{9} = 7$`
this is $$
 \color{blue}{c} = \color{blue}{\sqrt{7}}.
$$

---
class: clear, middle

---
# Example 10 (6/6)

- **vertices**: `$\color{darkorange}{V_1(5,2)}$` and `$\color{darkorange}{V_2(-1, 2)}$`. Here we use that `$\color{darkorange}{a}$` is the distance between the center `$\color{red}{C}(\color{green}{1}, \color{purple}{2})$` and a vertex.
 
 - **covertices**: `$\color{darkcyan}{B_1(1,5)}$` and `$\color{darkcyan}{B_2(1,-1)}$`. Here we use that `$\color{darkcyan}{b}$` is the distance between the center `$\color{red}{C}(\color{green}{1}, \color{purple}{2})$` and a covertex.

- **foci**: `$\color{blue}{F_1(1 + \sqrt{7} , 2) }$` and `$\color{blue}{F_2 (1 -\sqrt{7}, 2) }$`. Here we use that `$\color{blue}{c}$` is the distance between the center `$\color{red}{C}(\color{green}{1}, \color{purple}{2})$` and a focus.

---
# Example 11 (1/8)

An arch of a bridge is in the shape of a horizontal semiellipse. Its highest point is `$8$` meters, while its length is `$18$` meters.
 
 - Write an equation that represents this arch (the full ellipse).
 - How high is the arch at a point `$4$` meters away from the center horizontally?

**Solution** 
 
 Let's note that we are working with a horizontal (semi)ellipse, then the length in the statement is referred to the length of the major axis, and the _highest point_ corresponds to the distance between the center and a covertex of the ellipse.

---
# Example 11 (2/8)

We know:

- The **length of the major axis** `$\color{darkorange}{2a} = \color{darkorange}{18}$`, then `$\color{darkorange}{a} = \color{darkorange}{9}$`.
 
 - The value `$\color{darkcyan}{b} = \color{darkcyan}{8}$`.
 
 - The **position** of the ellipse is _horizontal_.

.center[
 **How do we get the center?**

Because any explicit point is given in the statement, we can **choose** that location of the center: let's take the origin `$\color{red}{C}(\color{green}{0}, \color{purple}{0})$`.
]

---
# Example 11 (3/8)

Then, using the previous information, the standard equation of the full ellipse is given by

$$
{\bf \frac{x^2}{\color{darkorange}{81}} + \frac{y^2}{ \color{darkcyan}{64}} = 1}.
$$

---
class: clear, middle

---
# Example 11 (5/8)

Now, because of the symmetry of the ellipse, to get the height of the arch at a point `$4$` meters away from the center horizontally, we just **consider a horizontal displacement to the right**.

- We don't know the height, so we represent it by `$\color{darkblue}{h}$`.
 - Now, in the graphical representation we consider the point `$\textbf{P}(4,\color{darkblue}{h})$`.

.center[
 **How do we get the value of** `$\color{darkblue}{h}$`?

]

---
# Example 11 (6/8)

.center[
 ATTENTION !!!
]

A point **lies on** a FIGURE if its coordinates **satisfy** the equation that REPRESENTS the given figure.

The previous sentence means that if we subtitute the coordinates of the point into the equation we obtain an equality.

---
# Example 11 (7/8)

Because the point `$\textbf{P}(4,\color{darkblue}{h})$` lies on the ellipse, after the subtitution into the standard equation, we have that $$
 \frac{4^2}{\color{darkorange}{81}} + \frac{\color{darkblue}{h}^2}{\color{darkcyan}{64}} = 1 .
$$

The problem request us to find the exact value of `$\color{darkblue}{h}$`, then we have to solve the previous equation for `$\color{darkblue}{h}$`, this is: $$
`\begin{gathered}
 \frac{\color{darkblue}{h}^2}{\color{darkcyan}{64}} = 1 - \frac{16}{\color{darkorange}{81}} \\
 \color{darkblue}{h}^2 = (64)\left(\frac{65}{81}\right) \\
 \color{darkblue}{h} = \pm \frac{8\sqrt{65}}{9}
\end{gathered}`
$$

---
# Example 11 (8/8)

Finally, because the height is a _distance_, we discard the negative solution for `$\color{darkblue}{h}$`. Then, the requested height is

$$
\color{darkblue}{h} = \frac{ 8 \sqrt{65} }{9}.
$$

---
# Example 12 (1/5)

According to Kepler's Laws, planets have elliptical orbits, with the Sun at one of the foci. The farthest Pluto gets from the Sun is `$7.4$` billions kilometers. The closest it gets to the Sun is `$4.4$` billions kilometers. Find an equation that represents the orbit of Pluto.

**Solution** 
 
 In order to solve this problem it is convenient to make a graphical representation of the information provided in the statement of the problem.

---
# Example 12 (2/5)

- Set `$\color{blue}{F}$` the position of the Sun (it is one focus).
 - Set `$\color{darkorange}{V_1}$` the farthest point Pluto gets from the Sun.
 - Set `$\color{darkorange}{V_2}$` the closest point Pluto gets to the Sun.
 - According to the statement, we have that $$
`\begin{gathered}
 d(\color{blue}{F}, \color{darkorange}{V_1}) = \color{#006400}{\bf (7.4)(10^{9})\text{ km}} \\
 d(\color{blue}{F}, \color{darkorange}{V_2}) = \color{#FF00FF}{\bf (4.4)(10^{9})\text{ km}}
\end{gathered}`
$$

---
# Example 12 (3/5)

- We note that `$\color{darkorange}{V_1}$` and `$\color{darkorange}{V_2}$` are the vertices of the ellipse, then the distance between them is the length of the major axis, and it is: $$
 \color{darkorange}{2a} = d(\color{darkorange}{V_1}, \color{darkorange}{V_2}) = d(\color{blue}{F}, \color{darkorange}{V_1}) + d(\color{blue}{F}, \color{darkorange}{V_2}) = \color{darkorange}{\bf (11.8)(10^9)}
 $$
this implies that $$
 \color{darkorange}{\bf a} = \color{darkorange}{\bf (5.9)(10^9)}
$$

- Because any explicit point is given in the statement, we can **choose** that location of the center: let's take the origin `$\color{red}{C}(\color{green}{0}, \color{purple}{0})$`.
 
---
# Example 12 (4/5)

- Next, to get `$\color{blue}{c} = d(\color{red}{C}, \color{blue}{F})$`, we note that $$
`\begin{aligned}
\color{blue}{c} & = d(\color{red}{C}, \color{darkorange}{V_2}) - d(\color{blue}{F}, \color{darkorange}{V_2}) \\
& = \color{darkorange}{ (5.9)(10^9) } - \color{#FF00FF}{ (4.4)(10^{9}) } \\
& = \color{blue}{(1.5)(10^9)}
\end{aligned}`
$$

- Let's find the value of `$\color{darkcyan}{b}^2$`: because `$\color{blue}{c}^2 = \color{darkorange}{a}^2 - \color{darkcyan}{b}^2$`, we obtain that $$
`\begin{aligned}
\color{darkcyan}{b}^2 & = \color{darkorange}{a}^2 - \color{blue}{c}^2 \\
& = ( \color{darkorange}{ (5.9)(10^9) } )^2 - ( \color{blue}{(1.5)(10^9)} )^2 \\
& = (34.81)(10^{18}) - (2.25)(10^{18}) \\
& = (32.56)(10^{18})
\end{aligned}`
$$

---
# Example 12 (5/5)

Finally, because `$\color{red}{C}(\color{green}{0}, \color{purple}{0})$`, `$\color{darkorange}{a}^2 = (34.81)(10^{18})$`, `$\color{darkcyan}{b}^2 = (32.56)(10^{18})$`, and our graphical representation considers that the major axis is **horizontal**, the standard equation that represents the orbit of Pluto is

$$
\frac{x^2}{(34.81)(10^{18})} + \frac{y^2}{(32.56)(10^{18})} = 1.
$$