Ellipse: Standard Equation

An ellipse is a conic section that is described as:

DEFINITION OF AN ELLIPSE:


The set of all points (x, y) in a plane the sum of whose distances from two fixed points, called foci, is constant.




Figure 1 shows a picture of an ellipse:



Thus for all (x, y), d1 + d2 = constant. When talking about an ellipse, the following terms are used:

  • The foci are two fixed points equidistant from the center of the ellipse.

  • The vertices are the points on the ellipse that fall on the line containing the foci.

  • The line segment or chord joining the vertices is the major axis.

  • The midpoint of the major axis is the center.

  • The axis perpendicular to the major axis is the minor axis.

Figure 1 shows an ellipse with a horizontal major axis. However an ellipse could also have a major axis that is vertical as shown in Figure 2.



The standard equation of an ellipse is:

STANDARD EQUATION OF AN ELLIPSE:


➢ Center coordinates (h, k)

➢ Major axis 2a

➢ Major axis 2b

➢ 0 < b < a


( xh ) 2 a 2 + ( yk ) 2 b 2 =1 major axis is horizontal


( xh ) 2 b 2 + ( yk ) 2 a 2 =1 major axis is vertical


The foci lie on the major axis, c units from the center, with c2 = a2 - b2.




Let's use these equations in some examples:

Example 1: Find the standard equation of the ellipse having foci at (2, 1) and (-4, 1) and a minor axis of 10.

Step 1: Determine the following:


The coordinates of the center (h, k).


The distance of half the minor axis (b).


The length of half the major axis (a).


The orientation of the major axis.

Center: Since the foci are equidistant from the center of the ellipse the center can be determine by finding the midpoint of the foci.


( h, k )=( 2+( 4 ) 2 ,  1+1 2 )=( 2 2 , 2 2 )=( 1,1 )


Length of b: The minor axis is given as 10, which is equal to 2b.


2b=10b=5


Length of a: To find a the equation c2 = a2 + b2 can be used but the value of c must be determined. Since c is the distance from the foci to the center, take either foci and determine the distance to the center. Then solve for a.


Foci (2, 1): c=| 2( 1 ) |=| 3 |=3


Foci (-4, 1): c=| 4 ( 1 ) |=| 3 |=3


c = 3


c 2 = a 2 b 2 a= c 2 + b 2


a= 3 2 + 5 2 a= 34


Orientation of major axis: Since the two foci fall on the horizontal line y = 1, the major axis is horizontal.

Step 2: Substitute the values for h, k, a and b into the equation for an ellipse with a horizontal major axis.

Horizontal major axis equation:


( xh ) 2 a 2 + ( yk ) 2 b 2 =1


Substitute values:


[ x( 1 ) ] 2 34 2 + ( y1 ) 2 5 2 =1


Simplify:


( x+1 ) 2 34 2 + ( y1 ) 2 5 2 =1

Example 2: Find the standard equation of an ellipse represented by x2 + 3y2 - 4x - 18y + 4 = 0.

Step 1: Group the x- and y-terms on the left-hand side of the equation.

x2 + 3y2 - 4x - 18y + 4 = 0


( x 2 4x)+(3 y 2 18y)+4=0

Step 2: Move the constant term to the right-hand side.

( x 2 4x)+(3 y 2 18y)=4

Step 3: Complete the square for the x- and y-groups.

( x 2 4x)+(3 y 2 18y)=4


Complete the square for the x-group


(x2 - 4x)


Take the coefficient of the x-term, divide by 2 and square the result.


4 2 =2; ( 2 ) 2 =4


Add the result to the x-group.


(x2 - 4x + 4)


Complete the square for the y-group


(3y2 - 18y)


Factor out a 3 so they y2-coefficient is 1


3(y2 - 6y)

Take the coefficient of the y-term, divide by 2 and square the result.


6 2 =3; ( 3 ) 2 =9


Add the result to the y-group.


3(y2 - 6y + 9)


Final result.


( x 2 4x+4)+3( y 2 6y+9)=4+?

Step 4: Add whatever was added to the left-hand side to the right-hand side.

( x 2 4x+4)+3( y 2 6y+9)=4+4+3( 9 )


( x 2 4x+4)+3( y 2 6y+9)=27

Step 5: Write the x-group and y-group as perfect squares.

( x2 ) 2 +3 ( y3 ) 2 =27

Step 6: Divide both sides by the value on the right-hand side.

( x2 ) 2 27 + 3 ( y3 ) 2 27 =1


( x2 ) 2 27 + ( y3 ) 2 9 =1





Related Links:
Math
algebra
Ellipse: Eccentricity
Hyperbola: Standard Equation
Pre Calculus


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