Algebra Examples
4x2+3y2=12
Step 1
Step 1.1
Divide each term by 12 to make the right side equal to one.
4x212+3y212=1212
Step 1.2
Simplify each term in the equation in order to set the right side equal to 1. The standard form of an ellipse or hyperbola requires the right side of the equation be 1.
x23+y24=1
x23+y24=1
Step 2
This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.
(x-h)2b2+(y-k)2a2=1
Step 3
Match the values in this ellipse to those of the standard form. The variable a represents the radius of the major axis of the ellipse, b represents the radius of the minor axis of the ellipse, h represents the x-offset from the origin, and k represents the y-offset from the origin.
a=2
b=√3
k=0
h=0
Step 4
The center of an ellipse follows the form of (h,k). Substitute in the values of h and k.
(0,0)
Step 5
Step 5.1
Find the distance from the center to a focus of the ellipse by using the following formula.
√a2-b2
Step 5.2
Substitute the values of a and b in the formula.
√(2)2-(√3)2
Step 5.3
Simplify.
Step 5.3.1
Raise 2 to the power of 2.
√4-(√3)2
Step 5.3.2
Rewrite √32 as 3.
Step 5.3.2.1
Use n√ax=axn to rewrite √3 as 312.
√4-(312)2
Step 5.3.2.2
Apply the power rule and multiply exponents, (am)n=amn.
√4-312⋅2
Step 5.3.2.3
Combine 12 and 2.
√4-322
Step 5.3.2.4
Cancel the common factor of 2.
Step 5.3.2.4.1
Cancel the common factor.
√4-322
Step 5.3.2.4.2
Rewrite the expression.
√4-31
√4-31
Step 5.3.2.5
Evaluate the exponent.
√4-1⋅3
√4-1⋅3
Step 5.3.3
Simplify the expression.
Step 5.3.3.1
Multiply -1 by 3.
√4-3
Step 5.3.3.2
Subtract 3 from 4.
√1
Step 5.3.3.3
Any root of 1 is 1.
1
1
1
1
Step 6
Step 6.1
The first vertex of an ellipse can be found by adding a to k.
(h,k+a)
Step 6.2
Substitute the known values of h, a, and k into the formula.
(0,0+2)
Step 6.3
Simplify.
(0,2)
Step 6.4
The second vertex of an ellipse can be found by subtracting a from k.
(h,k-a)
Step 6.5
Substitute the known values of h, a, and k into the formula.
(0,0-(2))
Step 6.6
Simplify.
(0,-2)
Step 6.7
Ellipses have two vertices.
Vertex1: (0,2)
Vertex2: (0,-2)
Vertex1: (0,2)
Vertex2: (0,-2)
Step 7
Step 7.1
The first focus of an ellipse can be found by adding c to k.
(h,k+c)
Step 7.2
Substitute the known values of h, c, and k into the formula.
(0,0+1)
Step 7.3
Simplify.
(0,1)
Step 7.4
The first focus of an ellipse can be found by subtracting c from k.
(h,k-c)
Step 7.5
Substitute the known values of h, c, and k into the formula.
(0,0-(1))
Step 7.6
Simplify.
(0,-1)
Step 7.7
Ellipses have two foci.
Focus1: (0,1)
Focus2: (0,-1)
Focus1: (0,1)
Focus2: (0,-1)
Step 8
Step 8.1
Find the eccentricity by using the following formula.
√a2-b2a
Step 8.2
Substitute the values of a and b into the formula.
√(2)2-(√3)22
Step 8.3
Simplify the numerator.
Step 8.3.1
Raise 2 to the power of 2.
√4-√322
Step 8.3.2
Rewrite √32 as 3.
Step 8.3.2.1
Use n√ax=axn to rewrite √3 as 312.
√4-(312)22
Step 8.3.2.2
Apply the power rule and multiply exponents, (am)n=amn.
√4-312⋅22
Step 8.3.2.3
Combine 12 and 2.
√4-3222
Step 8.3.2.4
Cancel the common factor of 2.
Step 8.3.2.4.1
Cancel the common factor.
√4-3222
Step 8.3.2.4.2
Rewrite the expression.
√4-312
√4-312
Step 8.3.2.5
Evaluate the exponent.
√4-1⋅32
√4-1⋅32
Step 8.3.3
Multiply -1 by 3.
√4-32
Step 8.3.4
Subtract 3 from 4.
√12
Step 8.3.5
Any root of 1 is 1.
12
12
12
Step 9
These values represent the important values for graphing and analyzing an ellipse.
Center: (0,0)
Vertex1: (0,2)
Vertex2: (0,-2)
Focus1: (0,1)
Focus2: (0,-1)
Eccentricity: 12
Step 10