# How to Graph an Ellipse Given an Equation

## What Is an Ellipse?

Ellipse is a locus of a point that moves such that the sum of its distances from two fixed points called foci is constant. The constant sum is the length of the major axis 2a.

**d _{1} + d_{2} = 2a**

Ellipse can also be defined as the locus of the point that moves such that the ratio of its distance from a fixed point called the focus, and a fixed-line called directrix, is constant and less than 1. The ratio of the distances may also be called as the eccentricity of the ellipse. Refer to the figure below.

**e = d _{3}/d_{4} < 1.0**

**e = c/a < 1.0**

### Properties and Elements of an Ellipse

**1. Pythagorean Identity**

a^{2} = b^{2} + c^{2}

**2. Length of Latus Rectum (LR)**

LR = 2b^{2} / a

**3. Eccentricity (First Eccentricity,e)**

e = c / a

**4. Distance from center to directrix (d)**

d = a / e

**5. Second Eccentricity (e')**

e' = c / b

**6. Angular Eccentricity (α)**

α = c / a

**7. Ellipse Flatness (f)**

f = (a - b) / a

**8. Ellipse Second Flatness (f')**

f' = (a - b) / b

**9. Area of an Ellipse (A)**

A = πab

**10. Perimeter of an Ellipse (P)**

P = 2π√ (a^{2} + b^{2})/2

### General Equation of an Ellipse

The general equation of an ellipse is where A ≠ C but have the same sign. The general equation of an ellipse is either of the following forms.

**Ax**^{2}**+ Cy**^{2}**+ Dx + Ey + F = 0****x**^{2}**+ Cy**^{2}**+ Dx + Ey + F = 0**

To solve for an ellipse, either one of the following conditions must be known.

1. Use general equation form when four (4) points along the ellipse are known.

2. Use the standard form when center (h,k) , semi-major axis a, and semi-minor axis b are known.

### Standard Equation of an Ellipse

The figure below shows the four (4) main standard equations for an ellipse depending on the location of the center (h,k). Figure 1 is the graph and standard equation for an ellipse with center at (0,0) of the cartesian coordinate system and the semi-major axis a lying along the x-axis. Figure 2 shows the graph and standard equation for an ellipse with center at (0,0) of the cartesian coordinate system and the semi-major axis a lies along the y-axis.

Figure 3 is the graph and standard equation for an ellipse with center at (h,k) of the cartesian coordinate system and the semi-major axis a parallel with the x-axis. Figure 4 shows the graph and standard equation for an ellipse with center at (h,k) of the cartesian coordinate system and the semi-major axis a parallel with the y-axis. The center (h,k) can be any point in the coordinate system.

Always take note that for an ellipse, semi-major axis a is always greater than semi-minor axis b. For an ellipse with a form Ax^{2} + Cy^{2} + Dx + Ey + F = 0, the center (h,k) can be obtained using the following formulas.

**h = - D / 2A**

**k = - E / 2C**

## Example 1

Given the general equation 16x^{2} + 25y^{2} - 128x - 150y + 381 = 0, graph the conic section and identify all important elements.

### Solution

a. Convert the general form to standard equation by completing the square. It is important to be knowledgeable with the process of completing the square in order to solve conic section problems like this. Then, solve for the coordinates of the center (h,k).

16x^{2} + 25y^{2} - 128x - 150y + 381 = 0

16x^{2} - 128x + ______ + 25y^{2} + 150y + ______ = - 381

16 (x^{2} - 8x + 16) + 25 (y^{2} - 6y +9) = - 381 + 256 +225

16 (x - 4)^{2} + 25 (y - 3)^{2} = 100^{}

[ (x - 4)^{2} / (25/4) ] + [ (y - 3)^{2} / 4 ] = 1 (*Standard form*)

**Center (h,k) = (4,3)**

b. Compute for the length of latus rectum (LR) using the formulas introduced earlier.

a^{2} = 25/4 and b^{2} = 4

a = 5/2 and b = 2

LR = 2b^{2} / a

LR = 2(2)^{2} / (5/2)

**LR = 3.2 units**

c. Compute for the distance (c) from the center (h,k) to focus.

a^{2} = b^{2} + c^{2}

(5/2)^{2} = (2)^{2} + c^{2}

**c = 3/2 units**

d1. Given the center (4,3), identify the coordinates of the focus and vertices.

**Right focus:**

F1_{x} = h + c

F1_{x} = 4 + 3/2

F1_{x} = 5.5

F1_{y} = k = 3

**F1 = (5.5 , 3)**

**Left focus:**

F2_{x} = h - c

F2_{x} = 4 - 3/2

F2_{x} = 2.5

F2_{y} = k = 3

**F2 = (2.5 , 3)**

d2. Given the center (4,3), identify the coordinates of the vertices.

**Right vertex:**

V1_{x} = h + a

V1_{x} = 4 + 5/2

V1_{x} = 6.5

V1_{y} = k = 3

**V1 = (6.5 , 3)**

**Left vertex:**

V2_{x} = h - a

V2_{x} = 4 - 5/2

V2_{x} = 1.5

V2_{y} = k = 3

**V2 = (1.5 , 3)**

e. Compute for the eccentricity of the ellipse.

e = c / a

e = (3/2) / (5/2)

**e = 3/5**

f. Solve for the distance of the directrix (d) from the center.

d = a / e

d = (5/2) / 0.6

**d = 25/6 units**

g. Solve for the area and perimeter of the ellipse given.

A = πab

A = π(5/2)(2)

**A = 5π square units**

P = 2π√ (a^{2} + b^{2})/2

P = 2π√ ((5/2)^{2} + 2^{2})/2

**P = 14.224 units**

## Example 2

Given the standard equation of an ellipse (x^{2 }/ 4) + (y^{2 }/ 16) = 1, identify the elements of the ellipse and graph the function.

### Solution

a. The given equation is already in standard form, so there is no need to complete the square. By method of observation, obtain the coordinates of the center (h,k).

(x^{2 }/ 4) + (y^{2 }/ 16) = 1

b^{2} = 4 and a^{2} = 16

a = 4

b = 2

**Center (h,k) = (0,0)**

b. Compute for the length of latus rectum (LR) using the formulas introduced earlier.

a^{2} = 16 and b^{2} = 4

a = 4 and b = 2

LR = 2b^{2} / a

LR = 2(2)^{2} / (4)

**LR = 2 units**

c. Compute for the distance (c) from the center (0,0) to focus.

a^{2} = b^{2} + c^{2}

(4)^{2} = (2)^{2} + c^{2}

**c = 2√3 units**

d1. Given the center (0,0), identify the coordinates of the focus and vertices.

**Upper focus:**

F1_{y} = k + c

F1_{y} = 0 + 2√3

F1_{y} = 2√3

F1_{x} = h = 0

**F1 = (0 , 2√3)**

**Lower focus:**

F2_{x} = k - c

F2_{x} = 0 - 2√3

F2_{x} = - 2√3

F2_{y} = h = 0

**F2 = (0 , - 2√3)**

d2. Given the center (0,0), identify the coordinates of the vertices.

**Upper vertex:**

V1_{y} = k + a

V1_{y }= 0 + 4

V1_{y} = 4

V1_{x} = h = 0

**V1 = (0 , 4)**

**Lower vertex:**

V2_{y} = k - a

V2_{y} = 0- 4

V2_{y} = - 4

V2_{x} = h = 0

**V2 = (0 , -4)**

e. Compute for the eccentricity of the ellipse.

e = c / a

e = (2√3) / (4)

**e = 0.866**

f. Solve for the distance of the directrix (d) from the center.

d = a / e

d = (4) / 0.866

**d = 4.62 units**

g. Solve for the area and perimeter of the ellipse given.

A = πab

A = π(4)(2)

**A = 8π square units**

P = 2π√ (a^{2} + b^{2})/2

P = 2π√ ((4)^{2} + 2^{2})/2

**P = 19.87 units**

## Example 3

The distance (center to center) of the moon from the earth varies from a minimum of 221,463 miles to a maximum of 252, 710 miles. Find the eccentricity of the moon's orbit.

### Solution

a. Solve for the semi-major axis "a".

2a = 221,463 + 252,710

**a = 237,086.5 miles**

b. Solve for the distance (c) of the earth from the center.

c = a - 221,463

c = 237,086.5 - 221,463

**c =15,623.5 miles**

c. Solve for the eccentricity.

e = c / a

e = 15,623.5 / 23,086.5

**e = 0.066**

## Did you learn from the examples?

## Learn How to Graph Other Conic Sections

- Graphing a Parabola in a Cartesian Coordinate System

The graph and location of a parabola depend on its equation. This is a step-by-step guide in graphing different forms of a parabola in the Cartesian coordinate system. - How to Graph a Circle Given a General or Standard Equation

Learn how to graph a circle given the general form and standard form. Familiarize with converting general form to standard form equation of a circle and know the formulas necessary in solving problems about circles.

## Questions & Answers

**© 2019 Ray**