How to Graph an Ellipse Given an Equation

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

Graphing an Ellipse Given an Equation
Graphing an Ellipse Given an Equation | Source

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

Definition of Ellipse
Definition of Ellipse | Source

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


Elements of an Ellipse
Elements of an Ellipse | Source

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

Standard Equations of Ellipse
Standard Equations of Ellipse | Source

Example 1

Given the general equation 16x2 + 25y2 - 128x - 150y + 381 = 0, graph the conic section and identify all important elements.

Graphing an Ellipse Given General Form of Equation
Graphing an Ellipse Given General Form of Equation | Source

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.

Graphing an Ellipse Given the Standard Form
Graphing an Ellipse Given the Standard Form | Source

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.

Graphing an Ellipse
Graphing an Ellipse | Source

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

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Questions & Answers

    © 2019 Ray

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        Isiaka Adio Abibu 

        7 days ago

        These exercises are quite educating

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