Table of Contents

Last modified on August 3rd, 2023

As we know, we can write an equation of an ellipse given its graph, we can also graph an ellipse given its equation. The equations help us to draw an ellipse on a graph with the x & y coordinate planes. This article will demonstrate how we can graph an ellipse from a given equation both in standard and general form.

To graph an ellipse, we first need to find out its center, foci, vertices, and co-vertices. The equations help us to find these parameters. From our ellipse article, we know that ellipses have 2 orientationsâ€” horizontal or vertical. The equations in both forms also help us identify its orientation.

Here we will see how an ellipse equation in standard form, centered at (h,k) is plotted on a graph.

The **standard form** of an equation for a **horizontal ellipse** (foci on major axis) **centered at (h,k)** is:

Â **(x – h)**^{2}**/a**^{2}** + (y – k)**^{2}**/b**^{2}** = 1, (a>b)**

Now, let us learn to plot an ellipse on a graph using an equation as in the above form.

Let’s take the equation (x – 1)^{2}/9 + (y – 2)^{2}/4= 1 and identify whether it is a horizontal or vertical ellipse. We will also label the center, vertices, co-vertices, and foci.

From the given equation, it is clear that the ellipse is horizontal, seeing the positions of a^{2} and b^{2} as in the denominators. So, the foci & vertices must be on the x-axis.

Comparing the given equation with standard form (x-h)^{2}/a^{2} + (y-k)^{2}/b^{2} = 1, we get:

Center: (h, k); h =1, k = 2, so Center: (1, 2)

Vertices: (h Â± a, k) = (4,2) & (-2,2)

Co-vertices: (h, k Â± b) = (1,4) & (1,0)

Length of Major Axis: 2a, Length of Minor Axis: 2b

âˆ´ a = 3, b = 2

Foci where F^{2} = a^{2} – b^{2} , F = (hÂ±F, k) {F = distance from center to the foci or focal length}

Equation of focal length F is F^{2} = a^{2} – b^{2} = 9-4 = 5, F = âˆš5 â‰ˆ 2.236

âˆ´ Foci = (1Â±âˆš5, 2) = (3.236,2) & (-1.236,2)

Now let’s graph the ellipse.

Now let’s solve some more examples when the ellipse equation is in standard form, not centered at the origin.

Graph the following ellipse : (x + 3)^{2}/9+ (y – 5)^{2}/3 = 1**. **Identify whether it is a horizontal or vertical ellipse and then find the vertices & foci.

Solution:

From the given equation:

(h,k) = (-3,5)

a = 3, b = âˆš3

vertices = (0,5) & (-6,5)

co-vertices = (âˆ’3,5Â±âˆš3) = (âˆ’3,6.73) & (âˆ’3,3.267)

F^{2} = a^{2 }– b^{2} , F = (hÂ±F, k) = (-3Â±âˆš3, 5) = (-1.27, 5) & (-4.73, 5)

Thus, it is a horizontal ellipse as found from the equation.

The **standard form** of an ellipse is for a **vertical** **ellipse** (foci on minor axis) **centered at (h,k)**

**(x – h)**^{2}**/b**^{2}** + (y – k)**^{2}**/a**^{2}** = 1 (a>b)**

Now, let us learn to plot an ellipse on a graph using an equation as in the above form.

Let’s take the equation x^{2}/25 + (y – 2)^{2}/36 = 1 and identify whether it is a horizontal or vertical ellipse. We will also label the center, vertices, co-vertices, and foci.

From the given equation, it is clear that the ellipse is vertical, seeing the positions of a^{2} and b^{2} as in the denominators. So, the foci & vertices must be on the y-axis.

Comparing the given equation with standard form (x-h)^{2}/b^{2} + (y-k)^{2}/a^{2} = 1, we get:

âˆ´ h =0, k = 2, Center (h, k) = (0,2)

Length of Major Axis: 2a, Length of Minor Axis: 2b

a = 6, b = 5

Vertices: (h, k Â± a) = (0,8) & (0,-4)

Co-vertices: (h Â± b, k) = (5,2) & (-5,2)

Equation of focal length F is F^{2} = a^{2} – b^{2} = 36-25 = 11, F = âˆš11 â‰ˆ 3.316 â‰ˆ 3.32

âˆ´ Foci = (h, kÂ±F) = (0, 2Â±âˆš11) = (0, 2Â±3.32) = (0,5.32) & (0,-1.32)

Now let’s graph the ellipse.

Graph the following ellipse: x^{2}+(y âˆ’ 1)^{2}/4=1. Find the foci.

Solution:

From the given equation, we get:

Center(h,k) = (0,1)

a = 2, b = 1

vertices = (0,3), (0,âˆ’1)

co-vertices = (1,1) & (âˆ’1,1)

F^{2} = a^{2} – b^{2} = 4-1, F = âˆš3

Foci = (h,kÂ±F) = (0,2.73) & (0,-0.73)

Here we will see how an ellipse equation in **general form**, **centered** at **(h,k)** is plotted on a graph. Ellipse having equation in general form is **px**^{2}** + qy**^{2}** ^{ }+ cx + dy + e = 0;** where

Now, let us learn to plot an ellipse on a graph using an equation as in the above form.

Let’s take the equation 9x^{2} + 4y^{2} + 18x – 16y -11 = 0 and identify whether it is a horizontal or vertical ellipse. We will also label the center, vertices, co-vertices, and foci.

First, let us rewrite the equation in the standard form.

9x^{2} + 4y^{2} + 18x – 16y -11 = 0

(p=9, q=4, c=18, d = -16, e = -11)

or, 9x^{2} + 18x + 4y^{2} + 16y = 11

or, 9(x^{2} + 2x) + 4(y^{2}-4x) = 11

Converting the expression to (aÂ±b)^{2}

we can write the expression as following:

9(x^{2} + 2x + __) + 4(y^{2}-4x + __) = 11

To get the values of __ in above expression,

for x, we divide the co-efficient of x (i.e. 2) by 2. We get 2/2 = 1

Similarly for y, we divide the co-efficient of y (i.e. 4) by 2. We get 4/2 = 2

So we write the expression as :

9(x^{2} + 2x + 1) + 4(y^{2}-4x + 4) = 11 + 9 + 16 (equating the two sides, we need to add 16 + 9 on both sides)

or, 9(x+1)^{2} + 4(y-2)^{2} = 36

or, [9(x+1)^{2} + 4(y-2)^{2}]/36 = 1

or, [(x+1)^{2}]/4+ [(y-2)^{2}]/9 = 1 (we get this expression as the equation of ellipse in standard form)

From this given equation, , it is clear that the ellipse is vertical, seeing the positions of a^{2} and b^{2} as in the denominators. So, the foci & vertices must be on the y-axis.

âˆ´ center (h, k) : h = -1, k = 2 Hence, center = (-1, 2)

a^{2} = 9; a = 3, b^{2} = 4; b = 2

so, the vertices are (-1, 5) & (-1, -1)

co-vertices are (1,2) & (-3,2)

F^{2} = a^{2} – b^{2} = 9-4 = 5 , F = âˆš5 â‰ˆ 2.236

Foci = (-1, 2Â± âˆš5) = (-1,4.236) & (-1, – 0.236)

Now let’s graph the ellipse.

Graph the ellipse 4x^{2} + 8x + 9y^{2} – 18y – 23 = 0

Solution:

From the given equation, simplifying it to standard form:

4x^{2} + 8x + 9y^{2} -18y = 23

or, 4(x^{2} + 2x + _) + 9 (y^{2}– 2y + _) = 23

or, 4(x^{2} + 2x + 1) + 9 (y^{2}– 2y + 1) = 23 +4 + 9 (adding +9 + 4 on both sides)

or, 4(x+1)^{2} + 9(y-1)^{2} = 36

or, (x+1)^{2}/9 + (y-1)^{2}/4 = 1 (we get standard form of equation of ellipse)

âˆ´ center (h,k) = (-1,1)

Vertices= (h Â± a, k) = (2,1) & (-4,1)

Co-vertices: (h, k Â± b) = (-1,3) & (-1,-1)

Here we will see how an ellipse equation in standard form, centered at the origin is plotted on a graph.

Let’s take the equation x^{2}/9 + y^{2}/1= 1 and identify whether it is a horizontal or vertical ellipse. We will also label the center, vertices, co-vertices, and foci.

Comparing the given equation in the question, with the standard equation of an ellipse x^{2}/a^{2} + y^{2}/b^{2} = 1centered at (0,0),

From the equation, we understand, the foci & vertices must be on the x-axis hence it is a horizontal ellipse, seeing the positions of a^{2} and b^{2} as in the denominators. So, the foci & vertices must be on the x-axis.

âˆ´Center = (0,0)

a= 3, b = 1

Vertices: (Â±a, 0) = (3,0) & (-3,0)

Co-vertices: (0, Â±b) = (0,1) & (0,-1)

Equation of focal length is F^{2} = a^{2} – b^{2} = 9-1 = 8, F = âˆš8 â‰ˆ 2.83

âˆ´ Foci = (Â±âˆš8, 0) = (2.83,0) & (-2.83,0)

Now let’s graph the ellipse.

Graph the ellipse x^{2} + Â y^{2}/25 = 1

Solution:

From the given equation, we get:

Center = (0,0)

a = 5, b = 1

Vertices = (0,5) & (0,-5)

Co-vertices = (1,0) & (-1,0)

Here we will see how an ellipse equation in general form, centered at the origin is plotted on a graph.

Ellipses centered at zero are represented in general form as **px**^{2}** + qy**^{2}** = e = 0**;where (p,q)>0

Now, let us learn to plot an ellipse on a graph using an equation as in the above form.

Let’s take the equation x^{2} + 16y^{2} -16= 0 and identify whether it is a horizontal or vertical ellipse. We will also label the center, vertices, co-vertices, and foci.

Rewriting the equation into standard form as follows:

x^{2} + 16y^{2} -16= 0

or, x^{2} + 16y^{2} = 16

or, x^{2}/16+ 16y^{2}/16 = 16/16 (dividing both sides by 16)

or, x^{2}/16 + y^{2}/1 = 1 (standard form)

From the equation, we understand, the foci & vertices must be on the x-axis hence it is a horizontal ellipse, seeing the positions of a^{2} and b^{2} as in the denominators. So, the foci & vertices must be on the x-axis.

âˆ´, center (h,k) = (0,0), a = 4, b = 1,

Vertices: (Â±a, 0) = (4,0) & (-4,0)

Co-vertices: (0, Â±b) = (0,1) & (0,-1)

F^{2} = 16-1 = 15, F = âˆš15 â‰ˆ 3.87 â‰ˆ 3.8

âˆ´ Foci = (Â±âˆš8, 0) = (3.8,0) & (-3.8,0)

Now let’s graph the ellipse.

Graph the ellipse 25x^{2} + 4y^{2} -100 = 0.

Solution:

From the given equation, writing the equation in standard form,

25x^{2} + 4y^{2} = 100

or, x^{2}/4 + y^{2}/25 = 1

âˆ´ we get: Center (h,k) = (0,0)

a = 5, b = 2.

Vertices = (0,5) & (0,-5)

Co-vertices = (2,0) & (-2,0)

Last modified on August 3rd, 2023