# How do you write an equation for the hyperbola with a centre at (-3,1), one focus (-5,1), length of conjugate axis 4?

Oct 16, 2016

One cannot have a hyperbola with the equation:

${\left(x - - 3\right)}^{2} / {0}^{2} - {\left(y - 1\right)}^{2} / {2}^{2} = 1$

and that is what your specifications imply.

#### Explanation:

Because the center and the given focus share the same y coordinate, we know that this is a horizontal transverse type with a general equation of the form:

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

We are given that the center $\left(h , k\right)$ is $\left(- 3 , 1\right)$ so we substitute these values into the equation:

${\left(x - - 3\right)}^{2} / {a}^{2} - {\left(y - 1\right)}^{2} / {b}^{2} = 1$

We are given the length of the conjugate axis is 4; this allows us to find b:

$2 b = 4$

$b = 2$

Substitute this into the equation:

${\left(x - - 3\right)}^{2} / {a}^{2} - {\left(y - 1\right)}^{2} / {2}^{2} = 1$

We are given that a focus is at $\left(- 5 , 1\right)$ and we know that the foci are at (h - c, k) and (h + c, k). Because c is always positive, we can only obtain a number more negative than the center's x coordinate, using the first focus:

$- 5 = h - c$

$- 5 = - 3 - c$

$- 2 = - c$

$c = 2$

We can find $a$ using the equation:

${c}^{2} = {a}^{2} + {b}^{2}$

${2}^{2} = {a}^{2} + {2}^{2}$

${a}^{2} = 0$

$a = 0$