Question

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

Answer 1

One cannot have a hyperbola with the equation:

`(x - -3)^2/0^2 - (y - 1)^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:

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

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

`(x - -3)^2/a^2 - (y - 1)^2/b^2 = 1`

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

`2b = 4`

`b = 2`

Substitute this into the equation:

`(x - -3)^2/a^2 - (y - 1)^2/2^2 = 1`

We are given that a focus is at `(-5, 1)` 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`