Graphing Hyperbolas
Key Questions
-
If It's known the equation of the hyperbolas, that is:
#(x-x_c)^2/a^2-(y-y_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)((x-h)^2/a^2-(y-k)^2/(a^2(e^2-1))=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{(x-1)^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,(-2-2)/2)=(1,-2)# - now finally to find
#e#
#ae# is the distance between the center and the focus#ae=3-1=2# #e=2# #color(blue)("Finally by substituting"# you get that the hyperbola's equation is as follows
#(x-1)^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)((y-k)^2/a^2-(x-h)^2/(a^2(e^2-1))=1# - the distance between the two vertices equals