Lesson Objectives
• Demonstrate an understanding of how to graph a circle
• Learn how to sketch the graph of an ellipse that is centered at the origin
• Learn how to sketch the graph of an ellipse that is shifted horizontally and vertically

## How to Graph an Ellipse

In the last lesson, we learned how to sketch the graph of a circle. A circle is a special case ellipse where the distance from the center to any point is the same. When we see the graph of an ellipse, it appears to be a squished circle. We will see two types of ellipses. The horizontal ellipse is short and fat, while the vertical ellipse is tall and skinny. In some textbooks, we will see separate formulas for each type of ellipse. In other textbooks, we will see one single formula. In general, an ellipse that is centered at the origin can be described using the following equations:

### Horizontal Ellipse (a > b)

$$\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$$ x-intercepts: (a,0), (-a,0)
y-intercepts: (0,b), (0,-b)

### Vertical Ellipse (a > b)

$$\frac{x^2}{b^2}+ \frac{y^2}{a^2}=1$$ x-intercepts: (b,0), (-b,0)
y-intercepts: (0,a), (0,-a)
Don't let yourself get confused over the two standard forms. In one case, a2 is the denominator for x2. This happens when the denominator underneath x2 is larger than the denominator underneath y2. Since x is the horizontal axis, this will stretch the graph horizontally, creating a short and fat ellipse. In the other case, a2 is the denominator for y2. This happens when the denominator underneath y2 is larger than the denominator underneath x2. Since y is the vertical axis, this will stretch the graph vertically, creating a tall and skinny ellipse. Either way, we see that a2 represents the larger denominator. When this is under x2, we have a horizontal ellipse and when this is under y2, we have a vertical ellipse.
To graph an ellipse that is centered at the origin, we plot the four intercepts and draw a smooth curve through the points. Let's look at an example.
Example 1: Sketch the graph of each ellipse. $$\frac{x^2}{4}+ \frac{y^2}{25}=1$$ Since 25, the number underneath y2 is larger than 4, the number underneath x2, we know we have a vertical ellipse. This means the graph will be tall and skinny. Following the model above, our intercepts will occur at:
x-intercepts: (2,0), (-2,0)
For the x-intercepts, we just take the principal and negative square root of the number underneath x2. This is how we get +2 and -2.
y-intercepts: (0,5), (0,-5)
For the y-intercepts, we just take the principal and negative square root of the number underneath y2. This is how we get +5 and -5.
To sketch the graph, we will plot the intercepts (2,0), (-2,0), (0,5), (0,-5), and then draw a smooth curve through the points.

### Graphing an Ellipse Shifted Horizontally and Vertically

When an ellipse is not centered at the origin, we can use the following equations:

### Horizontal Ellipse (a > b)

$$\frac{(x - h)^2}{a^2}+ \frac{(y-k)^2}{b^2}=1$$ This ellipse is shifted h units right and k units up from: $$\frac{x^2}{a^2}+ \frac{y}{b^2}=1$$ In other words, the center will now be at (h,k). Instead of finding the x-intercepts and the y-intercepts, we are just finding four points on the ellipse. We can do this by moving from the center (h,k) left a units, right a units, up b units, and down b units. In other words:
Center:
(h,k)
(h + a, k), (h - a, k), (h, k + b), (h, k - b)

### Vertical Ellipse (a > b)

$$\frac{(x - h)^2}{b^2}+ \frac{(y-k)^2}{a^2}=1$$ This ellipse is shifted h units right and k units up from: $$\frac{x^2}{b^2}+ \frac{y}{a^2}=1$$ For a vertical ellipse, the center will be found at (h,k).
Additional points will be found by moving from the center up and down by a units and then left and right by b units.
Center:
(h,k)
(h + b, k), (h - b, k), (h, k + a), (h, k - a)
Again, these two different formulas often cause a lot of confusion. Whatever is underneath the (x - h)2 is what we look at for determining horizontal movement. Whatever is underneath the (y - k)2 is what we look at for determining vertical movement. Just remember x is the horizontal axis and y is the vertical axis. Let's look at an example.
Example 2: Sketch the graph of each ellipse. $$\frac{(x - 4)^2}{9}+ \frac{(y + 1)^2}{16}=1$$ We can see that we have a vertical ellipse since the value of 16, which is underneath (y + 1)2 is larger than the value of 9, which is underneath (x - 4)2. Let's write our formula for a vertical ellipse: $$\frac{(x - h)^2}{b^2}+ \frac{(y-k)^2}{a^2}=1$$ Now we can rewrite the equation of our ellipse in this format: $$\frac{(x - 4)^2}{9}+ \frac{(y - (-1))^2}{16}=1$$ We know the center occurs at (h,k). Our center will occur at (4, -1). We will get two of our four points by moving left and right from the center by 3 units. We get this by taking the positive and negative square root of 9. This will give us the following two points:
(7,-1) and (1,-1)
Additionally, we can gain our other two points by moving up and down from the center by 4 units. We get this by taking the positive and negative square root of 16. This will give us two points:
(4,3) and (4,-5)
Now we can plot our four points and sketch the graph:

#### Skills Check:

Example #1

Find the center of the ellipse. $$\frac{(x - 8)^{2}}{144}+ \frac{(y + 3)^{2}}{64}=1$$

A
$$(-3, 8)$$
B
$$(3, -8)$$
C
$$(-8, 3)$$
D
$$(12, 8)$$
E
$$(8, -3)$$

Example #2

Find the x-intercepts. $$\frac{x^{2}}{81}+ \frac{y^{2}}{289}=1$$

A
$$(\sqrt{22}, 0), (-\sqrt{22}, 0)$$
B
$$(9, 0), (-9, 0)$$
C
$$(\sqrt{9}, 0), (-\sqrt{9}, 0)$$
D
$$(13, 0), (-13, 0)$$
E
$$(\sqrt{13}, 0), (-\sqrt{13}, 0)$$

Example #3

Find the y-intercepts. $$\frac{x^{2}}{3}+ \frac{y^{2}}{5}=1$$

A
$$(0, -3), (0, 3)$$
B
$$(0, -\sqrt{3}), (0, \sqrt{3})$$
C
$$(0, \sqrt{5}), (0, -\sqrt{5})$$
D
$$(0, 5), (0, -5)$$
E
$$(0, 8), (0, -8)$$