# How to Graph an Ellipse Given an Equation

Updated on December 31, 2019 Ray is a Licensed Civil Engineer and specializes in Structural Engineering. He loves to write anything about education.

## 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.

d1 + d2 = 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 = d3/d4 < 1.0

e = c/a < 1.0

### Properties and Elements of an Ellipse

1. Pythagorean Identity

a2 = b2 + c2

2. Length of Latus Rectum (LR)

LR = 2b2 / 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√ (a2 + b2)/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.

• Ax2 + Cy2 + Dx + Ey + F = 0
• x2 + Cy2 + 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 Ax2 + Cy2 + 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 16x2 + 25y2 - 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).

16x2 + 25y2 - 128x - 150y + 381 = 0

16x2 - 128x + ______ + 25y2 + 150y + ______ = - 381

16 (x2 - 8x + 16) + 25 (y2 - 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.

a2 = 25/4 and b2 = 4

a = 5/2 and b = 2

LR = 2b2 / a

LR = 2(2)2 / (5/2)

LR = 3.2 units

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

a2 = b2 + c2

(5/2)2 = (2)2 + c2

c = 3/2 units

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

Right focus:

F1x = h + c

F1x = 4 + 3/2

F1x = 5.5

F1y = k = 3

F1 = (5.5 , 3)

Left focus:

F2x = h - c

F2x = 4 - 3/2

F2x = 2.5

F2y = k = 3

F2 = (2.5 , 3)

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

Right vertex:

V1x = h + a

V1x = 4 + 5/2

V1x = 6.5

V1y = k = 3

V1 = (6.5 , 3)

Left vertex:

V2x = h - a

V2x = 4 - 5/2

V2x = 1.5

V2y = 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√ (a2 + b2)/2

P = 2π√ ((5/2)2 + 22)/2

P = 14.224 units

## Example 2

Given the standard equation of an ellipse (x2 / 4) + (y2 / 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).

(x2 / 4) + (y2 / 16) = 1

b2 = 4 and a2 = 16

a = 4

b = 2

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

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

a2 = 16 and b2 = 4

a = 4 and b = 2

LR = 2b2 / a

LR = 2(2)2 / (4)

LR = 2 units

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

a2 = b2 + c2

(4)2 = (2)2 + c2

c = 2√3 units

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

Upper focus:

F1y = k + c

F1y = 0 + 2√3

F1y = 2√3

F1x = h = 0

F1 = (0 , 2√3)

Lower focus:

F2x = k - c

F2x = 0 - 2√3

F2x = - 2√3

F2y = h = 0

F2 = (0 , -2√3)

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

Upper vertex:

V1y = k + a

V1y = 0 + 4

V1y = 4

V1x = h = 0

V1 = (0 , 4)

Lower vertex:

V2y = k - a

V2y = 0- 4

V2y = - 4

V2x = 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√ (a2 + b2)/2

P = 2π√ ((4)2 + 22)/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

See results