# 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!

Jun 28, 2017

#### Explanation:

None your parabolas are going to fit the standard forms:

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1$

because they are rotated hyperbolas.

The only equation that they are going to fit is equation $\text{(9.4.1)}$ in the reference:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0 \text{ (9.4.1)}$

For example, equation [1]:

$y = \frac{x + 3}{x - 1} \text{ [1]}$

can be made to fit (9.4.1) by multiplying both sides by $\left(x - 1\right)$

$x y - y = x + 3$

and then moving everything to the left side:

$x y - x - y - 3 = 0 \text{ [1.1]}$

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

How to sketch the equation $y = \frac{x + 3}{x - 1} \text{ [1]}$:

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 $\left(1 , 1\right)$
2. Please observe that in equation [1] the numerator forces a $y = 0$ condition at the point $x = - 3$. This gives you the point $\left(- 3 , 0\right)$.
3. Because the hyperbola is rotated $\frac{\pi}{4}$ the vertices will be on the line $y = x$; they are the points $\left(- 1 , - 1\right)$ and $\left(3 , 3\right)$

Here is a graph of $y = \frac{x + 3}{x - 1} \text{ [1]}$, 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.