How do you graph a hyperbola when given the equation in standard form,

*Ax*^{2} + *Bxy* + *Cy*^{2} +* Dx* + *Ey* + *F* = 0 instead of graph form?

You will need to transform the equation in order to identify the center point and the radii in order to sketch the graph.

Transformation involves completing the square as for a circle with a couple of extra steps.

4*x*^{2} - 9*y*^{2} - 16*x* + 90*y* - 245= 0

Given this equation in standard form where A = 4, B = 0, C = -9, D = -16, E = 10, F = 205, notice that the y^{2} term is negative, which determines this is a hyperbola.

4*x*^{2} - 16*x* - 9*y*^{2} + 90*y* = 245

Rearrange to group like terms and move constant to opposite of equation.

4(*x*^{2} - 4*x* + ___) - 9(*y*^{2} - 10*y* + ___) = 245

Group “x” terms and “y” terms and factor out the coefficients of the squared terms. Notice factoring out the -9 makes the 10 negative as well.

4(*x*^{2} - 4*x* + ___)- 9(*y*^{2} - 10*y* + 25) = 245

+ (4 x 4) + (-9 x 25)

Complete each of the "squares" by adding the appropriate quantities and add like quantities to the opposite side of the equation.

4(x - 2)^{2} - 9(y - 5)^{2} = 36

Write perfect squares in factored form and combinethe constants.

four times x minus two squared over thirty-six
4(*x* − 2)^{2}
36
−
nine times y minus five squared over thirty-six
9(*y* − 5)^{2}
36
=
thirty-six over thirty-six
36
36

Divide each term by 36 so the left side equals 1.

x minus two squared over nine
(*x* − 2)^{2}
9
−
y minus five squared over four
(*y* − 5)^{2}
4
= 1

Simplify all terms!

Can you now identify the following critical attributes of the hyperbola?

Use your own graph paper or go to Print Free Graph Paper and sketch the graph.

*h* = __________ *k* = __________ center point __________

*horizontal*-radius: _________ *vertical* radius:__________

Interactive popup. Assistance may be required.
*Horizontal*-radius is 3 units

*Vertical*-radius is 2 units

Hyperbola opens in the _________ direction