In the previous chapter, Conic Sections: Ellipses, I have discussed about what an Ellipse is and its important properties. In this lesson, I will go through the mathematical equations of ellipses.

The equation of an ellipse depends on its orientation and center. We’ll explore the two main cases: 1. Ellipse Centered at the Origin (0, 0) & 2. Ellipse Centered at (h, k) .

1. Ellipse Centered at the Origin (0, 0)

a) Major Axis along the x-axis:

The standard equation is:

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

Where:

• a is the semi-major axis (half the length of the major axis).
• b is the semi-minor axis (half the length of the minor axis).
• The vertices are at (±a, 0).
• The co-vertices are at (0, ±b).
• The foci are at (±c, 0), where ##c = \sqrt{a^2 – b^2}##.

Example 1:

Find the equation of an ellipse with vertices at (±5, 0) and foci at (±3, 0).

Solution:

Since the vertices are at (±5, 0), we know that a = 5. The foci are at (±3, 0), so c = 3. We can find b using the relationship ##c^2 = a^2 – b^2##:

### 3^2 = 5^2 – b^2 ###### 9 = 25 – b^2 ###### b^2 = 16 ###### b = 4 ###

Therefore, the equation of the ellipse is:

### \frac{x^2}{25} + \frac{y^2}{16} = 1 ###

b) Major Axis along the y-axis:

The standard equation is:

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

Where:

• a is the semi-major axis.
• b is the semi-minor axis.
• The vertices are at (0, ±a).
• The co-vertices are at (±b, 0).
• The foci are at (0, ±c), where ##c = \sqrt{a^2 – b^2}##.

Example 2:

Find the equation of an ellipse with vertices at (0, ±7) and foci at (0, ±2).

Solution:

Since the vertices are at (0, ±7), we know that a = 7. The foci are at (0, ±2), so c = 2. We can find b using the relationship ##c^2 = a^2 – b^2##:

### 2^2 = 7^2 – b^2 ###### 4 = 49 – b^2 ###### b^2 = 45 ###### b = \sqrt{45} ###

Therefore, the equation of the ellipse is:

### \frac{x^2}{45} + \frac{y^2}{49} = 1 ###

2. Ellipse Centered at (h, k)

When the center of the ellipse is at a point (h, k), the equations are shifted accordingly.

a) Major Axis Parallel to the x-axis:

The standard equation is:

### \frac{(x – h)^2}{a^2} + \frac{(y – k)^2}{b^2} = 1 ###

Where:

• The center is at (h, k).
• The vertices are at (h ± a, k).
• The co-vertices are at (h, k ± b).
• The foci are at (h ± c, k), where ##c = \sqrt{a^2 – b^2}##.

Example 3:

Find the equation of an ellipse with center at (2, -1), a semi-major axis of 4 along the x-axis, and a semi-minor axis of 3.

Solution:

We are given that h = 2, k = -1, a = 4, and b = 3. Therefore, the equation is:

### \frac{(x – 2)^2}{16} + \frac{(y + 1)^2}{9} = 1 ###

b) Major Axis Parallel to the y-axis:

The standard equation is:

### \frac{(x – h)^2}{b^2} + \frac{(y – k)^2}{a^2} = 1 ###

Where:

• The center is at (h, k).
• The vertices are at (h, k ± a).
• The co-vertices are at (h ± b, k).
• The foci are at (h, k ± c), where ##c = \sqrt{a^2 – b^2}##.

Example 4:

Find the equation of an ellipse with center at (-3, 4), a semi-major axis of 6 along the y-axis, and a semi-minor axis of 2.

Solution:

We are given that h = -3, k = 4, a = 6, and b = 2. Therefore, the equation is:

### \frac{(x + 3)^2}{4} + \frac{(y – 4)^2}{36} = 1 ###

Graphing Ellipses

Graphing an ellipse involves several steps, depending on the equation you are given. Here’s a general approach:

1. Identify the Center: Determine the center (h, k) from the equation.
2. Determine the Orientation: Check which denominator is larger (a² or b²). If a² is under the x-term, the major axis is horizontal. If a² is under the y-term, the major axis is vertical.
3. Find the Vertices: Calculate the vertices using the center and the value of a.
4. Find the Co-vertices: Calculate the co-vertices using the center and the value of b.
5. Find the Foci: Calculate the foci using the center and the value of c (where ##c = \sqrt{a^2 – b^2}##).
6. Plot the Points: Plot the center, vertices, co-vertices, and foci on the coordinate plane.
7. Sketch the Ellipse: Draw a smooth curve through the vertices and co-vertices to form the ellipse.

Example 5:

Graph the ellipse: ##\frac{(x – 1)^2}{9} + \frac{(y + 2)^2}{4} = 1##

Solution:

1. Center: The center is at (1, -2).
2. Orientation: Since 9 is under the x-term, the major axis is horizontal.
3. Vertices: a² = 9, so a = 3. The vertices are at (1 ± 3, -2), which are (4, -2) and (-2, -2).
4. Co-vertices: b² = 4, so b = 2. The co-vertices are at (1, -2 ± 2), which are (1, 0) and (1, -4).
5. Foci: ##c = \sqrt{a^2 – b^2} = \sqrt{9 – 4} = \sqrt{5}##. The foci are at (1 ± √5, -2).
6. Plot and Sketch: Plot the center, vertices, co-vertices, and foci. Then, sketch the ellipse.

Applications of Ellipses

Ellipses have numerous real-world applications. Here are a few examples:

• Astronomy: The orbits of planets around the sun are elliptical, with the sun at one focus.
• Engineering: Elliptical gears are used in various machines to convert rotational motion.
• Architecture: Elliptical arches are used in the construction of bridges and buildings for their strength and aesthetic appeal.
• Acoustics: Elliptical rooms have a unique property: a sound emitted from one focus will reflect off the walls and converge at the other focus. This is known as the “whispering gallery” effect.
• Medicine: Lithotripsy, a medical procedure to break up kidney stones, uses an elliptical reflector to focus sound waves on the stone.

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