Graphing Hyperbolas
Key Questions

If It's known the equation of the hyperbolas, that is:
#(xx_c)^2/a^2(yy_c)^2/b^2=+1# ,we can graph the hyperbolas in this way:

find the center
#C(x_c,y_c)# ; 
make a rectangle with the center in
#C# and with sides#2a# and#2b# ; 
draw the lines that pass from the opposite vertices of the rectangle (the asymptotes);

if the sign of
#1# is#+# , than the two branches are left and right of the rectangule and the vertices are in the middle of the vertical sides, if the sign of#1# is## , than the two branches are up and down of the rectangule and the vertices are in the middle of the horizontal sides.


Answer:
check the explanation below.
Explanation:
#Ax^2+By^2+Cxy+Dx+Ey+F=0# That's the general equation of any conic section including the hyperbola
where you can find the equation of a hyperbola given enough points
where
#color(green)(" A,B,C,D,E,F# are constants
#color(red)("where either A or B are negative but never both"# though this is usually too hard to solve this type of equation manually
#color(blue)("Special Case:"#
the hyperbola is horizontal or verticalif it's horizontal you could use this
#color(green)((xh)^2/a^2(yk)^2/(a^2(e^21))=1#
#color(green)("where a is a constant and "e" is the eccentricity "(h,k)" is the center of the hyperbola"# #color(blue)("Example :")# graph{(x1)^2(y+2)^2/3=1 [3.502, 5.262, 3.852, 0.533]}
given one of it's foci points is#(3,2)# Using the properties of the hyperbola to determine the constants
 the distance between the two vertices equals
#2a#
#2a=2# #rarr# #a=1#  now to get the center
#(h,k)#
it can simply be done by finding the midpoint of the line segment joining the two vertices
#((0+2)/2,(22)/2)=(1,2)#  now finally to find
#e#
#ae# is the distance between the center and the focus#ae=31=2# #e=2# #color(blue)("Finally by substituting"# you get that the hyperbola's equation is as follows
#(x1)^2/1(y+2)^2/2=1# #color(green)("Additional tricks that could help :)"# distance between the directrix and the center of the hyperbola
#color(green)(a/e# if it's vertical the equation will change but with the same solving technique and it will be
#color(green)((yk)^2/a^2(xh)^2/(a^2(e^21))=1#  the distance between the two vertices equals