__Category__: Analytic Geometry, Algebra"Published in Newark, California, USA"

Sketch the graph for

__Solution__:

Consider the given equation above

The above equation represents a hyperbola because one of the 2nd degree variable is negative. If y

^{2}is negative, then the curve of the hyperbola opens to the left and to the right. If x

^{2}is negative, then the curve of the hyperbola opens upward and downward.

The first step is to reduce the given equation of the hyperbola to its standard form as follows

Group the above terms according to its variables

Complete the square of the grouped terms

Divide both sides of the equation by 36

Since y is negative, then the curve of the hyperbola opens to the left and to the right. The transverse axis is parallel to the x-axis where the center, two foci, and two vertices are located.

The center of the hyperbola is

**C(-1, 2)**.

The semi-major axis of the hyperbola is

Therefore, the main vertices of the hyperbola are

V

_{1}(-1 + 2, 2) =

**V**

_{1}(1, 2)V

_{2}(-1 - 2, 2) =

**V**

_{2}(-3, 2)The semi-minor axis of the hyperbola is

Therefore, the other vertices of the hyperbola are

V

_{3}(-1, 2 + 1½) = V

_{3}(-1, 3½)

V

_{4}(-1, 2 - 1½) = V

_{4}(-1, ½)

These two vertices of the hyperbola will not be used to draw the curve proper but it is helpful to draw the imaginary rectangle. To draw the imaginary rectangle, draw a light vertical line that passes through V

_{1}, another light vertical line that passes through V

_{2}, a light horizontal line that passes through V

_{3}, and another light horizontal line that passes through V

_{4}. The resulting imaginary rectangle will be a guide to draw the two asymptotes of the hyperbola in which you will connect and extend the two opposite vertices of a rectangle. The two asymptotes of the hyperbola should meet or intersect at the center of the hyperbola. The two equations of the asymptotes of the hyperbola are

Change the right side of the equation to zero, we have

Factor the above equation as the sum and the difference of two squares

Equate each factor to zero to get the equations of the two asymptotes.

For the first asymptote of the hyperbola

Multiply both sides of the equation by 6, we have

For the second asymptote of the hyperbola

Multiply both sides of the equation by 6, we have

Next, we need to get the coordinates of the foci and the ends of the latera recta of the hyperbola. First, we need to solve for the value of c which is the distance of the focus to the center of the hyperbola as follows

Therefore, the two foci of the hyperbola are

F

_{1}(-1 + 2½, 2) =

**F**

_{1}(1½, 2)F

_{2}(-1 - 2½, 2) =

**F**

_{2}(-3½, 2)Next, solve for the length of the latus rectum as follows

Therefore, the end points of the latera recta are

L

_{1}(1½, 2 + 1⅛) =

**L**

_{1}(**1½, 3⅛)**

L

_{2}(1½, 2 - 1⅛) =

**L**

_{2}(**1½, ⅞)**

L

_{3}(-3½, 2 + 1⅛) =

**L**

_{3}(**-3½, 3⅛)**

L

_{4}(-3½, 2 - 1⅛) =

**L**

_{4}(**-3½, ⅞)**

Finally, connect the points L

_{3}, V

_{2}, and L

_{4}to draw a curve that opens to the left and connect the points L

_{1}, V

_{1}, and L

_{2}to draw a curve that opens to the right as follows

Photo by Math Principles in Everyday Life |