Question

How sketch complicated hyperbola?

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

Answer 1

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:

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.