How sketch complicated hyperbola?

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Can somebody please explain to me how make it into the general hyperbola form for any of those questions? Thanks a lot!

1 Answer
Jun 28, 2017

Answer:

Please read Rotation of Axes

Explanation:

None your parabolas are going to fit the standard forms:

#(x-h)^2/a^2-(y-k)^2/b^2=1#

#(y-k)^2/a^2-(x-h)^2/b^2=1#

because they are rotated hyperbolas.

The only equation that they are going to fit is equation #"(9.4.1)"# in the reference:

#Ax^2+Bxy+Cy^2+Dx+Ey+F = 0" (9.4.1)"#

For example, equation [1]:

#y = (x+3)/(x-1)" [1]"#

can be made to fit (9.4.1) by multiplying both sides by #(x-1)#

#xy-y = x+3#

and then moving everything to the left side:

#xy -x -y -3=0" [1.1]"#

Please observe that equation [1.1] is equation (9.4.1) with #A = C = 0, B =1, D=E=-1, and E = -3#. If you read the comment following equation (9.4.6), you will see that all of your hyperbolas are rotated by #pi/4"radians"#

How to sketch the equation #y = (x+3)/(x-1)" [1]"#:

  1. Please observe that denominator in equation [1] forces a divide by zero condition at #x =1#. This means that the asymptotes are the lines #x = 1# and #y = 1# and the center of the hyperbola is the intersection of the asymptotes, the point #(1,1)#
  2. Please observe that in equation [1] the numerator forces a #y = 0# condition at the point #x =-3#. This gives you the point #(-3,0)#.
  3. Because the hyperbola is rotated #pi/4# the vertices will be on the line #y = x#; they are the points #(-1,-1)# and #(3,3)#

Here is a graph of #y = (x+3)/(x-1)" [1]"#, with the asymptotes and the points:

Desmos.com

Both of your remaining equations can be sketched in exactly the same way. Please give it a try and feel free to ask questions, if you have any.