(x-9)29-(y-10)27=1

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.

(x-9)29-(y-10)27=1

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

(x-h)2a2-(y-k)2b2=1

Match the values in this hyperbola to those of the standard form. The variable h represents the x-offset from the origin, k represents the y-offset from origin, a.

a=3

b=7

k=10

h=9

The center of a hyperbola follows the form of (h,k). Substitute in the values of h and k.

(9,10)

Find the distance from the center to a focus of the hyperbola by using the following formula.

a2+b2

Substitute the values of a and b in the formula.

(3)2+(7)2

Simplify.

Raise 3 to the power of 2.

9+(7)2

Rewrite 72 as 7.

Use axn=axn to rewrite 7 as 712.

9+(712)2

Apply the power rule and multiply exponents, (am)n=amn.

9+712⋅2

Combine 12 and 2.

9+722

Cancel the common factor of 2.

Cancel the common factor.

9+722

Divide 1 by 1.

9+71

9+71

Evaluate the exponent.

9+7

9+7

Simplify the expression.

Add 9 and 7.

16

Rewrite 16 as 42.

42

Pull terms out from under the radical, assuming positive real numbers.

4

4

4

4

The first vertex of a hyperbola can be found by adding a to h.

(h+a,k)

Substitute the known values of h, a, and k into the formula and simplify.

(12,10)

The second vertex of a hyperbola can be found by subtracting a from h.

(h-a,k)

Substitute the known values of h, a, and k into the formula and simplify.

(6,10)

The vertices of a hyperbola follow the form of (h±a,k). Hyperbolas have two vertices.

(12,10),(6,10)

(12,10),(6,10)

The first focus of a hyperbola can be found by adding c to h.

(h+c,k)

Substitute the known values of h, c, and k into the formula and simplify.

(13,10)

The second focus of a hyperbola can be found by subtracting c from h.

(h-c,k)

Substitute the known values of h, c, and k into the formula and simplify.

(5,10)

The foci of a hyperbola follow the form of (h±a2+b2,k). Hyperbolas have two foci.

(13,10),(5,10)

(13,10),(5,10)

Find the eccentricity by using the following formula.

a2+b2a

Substitute the values of a and b into the formula.

(3)2+(7)23

Simplify the numerator.

Raise 3 to the power of 2.

9+(7)23

Rewrite 72 as 7.

Use axn=axn to rewrite 7 as 712.

9+(712)23

Apply the power rule and multiply exponents, (am)n=amn.

9+712⋅23

Combine 12 and 2.

9+7223

Cancel the common factor of 2.

Cancel the common factor.

9+7223

Divide 1 by 1.

9+713

9+713

Evaluate the exponent.

9+73

9+73

Add 9 and 7.

163

Rewrite 16 as 42.

423

Pull terms out from under the radical, assuming positive real numbers.

43

43

43

Find the value of the focal parameter of the hyperbola by using the following formula.

b2a2+b2

Substitute the values of b and a2+b2 in the formula.

724

Rewrite 72 as 7.

Use axn=axn to rewrite 7 as 712.

(712)24

Apply the power rule and multiply exponents, (am)n=amn.

712⋅24

Combine 12 and 2.

7224

Cancel the common factor of 2.

Cancel the common factor.

7224

Divide 1 by 1.

714

714

Evaluate the exponent.

74

74

74

The asymptotes follow the form y=±b(x-h)a+k because this hyperbola opens left and right.

y=±73⋅(x-(9))+10

Remove parentheses.

y=73⋅(x-(9))+10

Simplify each term.

Multiply -1 by 9.

y=73⋅(x-9)+10

Apply the distributive property.

y=73x+73⋅-9+10

Combine 73 and x.

y=7×3+73⋅-9+10

Cancel the common factor of 3.

Factor 3 out of -9.

y=7×3+73⋅(3(-3))+10

Cancel the common factor.

y=7×3+73⋅(3⋅-3)+10

Rewrite the expression.

y=7×3+7⋅-3+10

y=7×3+7⋅-3+10

Move -3 to the left of 7.

y=7×3-37+10

y=7×3-37+10

y=7×3-37+10

Remove parentheses.

y=-73⋅(x-(9))+10

Simplify each term.

Multiply -1 by 9.

y=-73⋅(x-9)+10

Apply the distributive property.

y=-73x-73⋅-9+10

Combine x and 73.

y=-x73-73⋅-9+10

Cancel the common factor of 3.

Move the leading negative in -73 into the numerator.

y=-x73+-73⋅-9+10

Factor 3 out of -9.

y=-x73+-73⋅(3(-3))+10

Cancel the common factor.

y=-x73+-73⋅(3⋅-3)+10

Rewrite the expression.

y=-x73-7⋅-3+10

y=-x73-7⋅-3+10

Multiply -3 by -1.

y=-x73+37+10

y=-x73+37+10

y=-x73+37+10

This hyperbola has two asymptotes.

y=7×3-37+10,y=-x73+37+10

These values represent the important values for graphing and analyzing a hyperbola.

Center: (9,10)

Vertices: (12,10),(6,10)

Foci: (13,10),(5,10)

Eccentricity: 43

Focal Parameter: 74

Asymptotes: y=7×3-37+10, y=-x73+37+10

Graph ((x-9)^2)/9-((y-10)^2)/7=1