Find the vertices and foci of the hyperbola with equation (x-5)^2/144 - (y-4)^2/81 = 1.

Question

anonymous · Answer

Vertices: (14, 4), (-4, 4); Foci: (-4, 4), (14, 4)

 Vertices: (17, 4), (-7, 4); Foci: (-10, 4), (20, 4)

 Vertices: (4, 17), (4, -7); Foci: (4, -10), (4, 20)

 Vertices: (4, 14), (4, -4); Foci: (4, -4), (4, 14)

e.mccormick · Answer

Do you know how to find any of those?

anonymous · Answer

no i would like to know how to solve the problem

e.mccormick · Answer

OK.  The first thing to note is what is being subtracted includes a y.  That means the hyperbola is basically this shape:

|dw:1401839625678:dw|

e.mccormick · Answer

The standard form is:

$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$

Now, there is also a c involved.

$a^2+b^2=c^2$

|dw:1401839880320:dw|

The b values are used to find the asymptotes.