# FIND THE LENGTH OF THE MAJOR OR MINOR AXES OF AN ELLIPSE

## About "Find the length of the major or minor axes of an ellipse"

Find the length of the major or minor axes of an ellipse :

The formula to find the length of major and minor axes are always same, if its center is (0, 0) or not.

Major axis :

The line segment AA′ is called the major axis and the length of the major axis is 2a.

Minor axis :

The line segment BB′ is called the major axis and the length of the major axis is 2b. ## Find the length of the major or minor axes of an ellipse - Examples

Let us see some example problems based on the above concept.

Example 1 :

Find the length of major and minor axes of the following ellipse.

x²/9 + y²/4  =  1

Solution :

To find the length of major and minor axis, first we have to find the length of a and b. Here the greatest value is known as "a²" and smallest value is known as "b²".

a² = 9 and b² = 4. Since the denominator of the variable y is greater, the ellipse is symmetric about y-axis.

So, a = 3 and b = 2

length of major axis  =  2a ==> 2(3)  = 6 units

length of minor axis  =  2b ==> 2(2) = 4 units

Example 2 :

Find the length of major and minor axes of the following ellipse.

36x² + 4y² - 72x + 32y - 44 = 0

Solution :

Since the given equation is in the standard form, we have to make it is as general form.

36x² - 72x + 4y² + 32y - 44 = 0

36(x² - 2x) + 4(y² + 8y) - 44 = 0

36 [x²-2x(1)+1²-1²] + 4 [y²+ 2y(4)+4²-4²] - 44 = 0

36 [(x - 1)²- 1] + 4[(y - 4)²- 16] - 44  =  0

36 (x - 1)²- 36 + 4(y - 4)²- 64 - 44  =  0

36 (x - 1)²+ 4(y - 4)² - 36 - 64 - 44  =  0

36 (x - 1)²+ 4(y - 4)² = 144

Divide the whole equation by 144

(x - 1)²/4 + (y - 4)²/36 = 1

The ellipse is symmetric about y-axis.

a² = 36 , b² = 4

a = 6 and b = 2

length of major axis  =  2a ==> 2(6)  = 12 units

length of minor axis  =  2b ==> 2(2) = 4 units

Example 3 :

Find the length of major and minor axes of the following ellipse.

4x² + 3y² + 8x + 12y + 4 = 0

Solution :

Since the given equation is in the standard form, we have to make it is as general form.

4x² + 8x + 3y²+ 12y + 4 = 0

4(x² + 2x) + 3(y² + 4y) + 4 = 0

4 [x²+2x(1)+1²-1²] + 3 [y²+ 2y(2)+2²-2²] + 4 = 0

4 [(x + 1)² - 1²] + 3 [(y + 2)²-2²] + 4 = 0

4 [(x + 1)² - 1] + 3 [(y + 2)²-4] + 4 = 0

4 (x + 1)² - 4 + 3 (y + 2)²- 12 + 4 = 0

4 (x + 1)² + 3 (y + 2)²- 12 + 4 - 4 = 0

4 (x + 1)² + 3 (y + 2)²- 12 = 0

4 (x + 1)² + 3 (y + 2)² = 12

Divide the whole equation by 12

(x - 1)²/3 + (y - 4)²/4 = 1

The ellipse is symmetric about y-axis.

a² = 4 , b² = 3

a = 2 and b = √3

length of major axis  =  2a ==> 2(2)  = 4 units

length of minor axis  =  2b ==> 2(√3) = 2√3 units

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