find the equation of the family of hyperbolas where foci are (~4, 6) & (4, 0). Find the member of the family that passes through (2, 0)

Answer 1

I will assume that you meant:
A(-4,6) and B(4,0) for the foci

Let P(x,y) be any point on the hyperbola
By the basic definition of a hyperbola:
|PA - PB| = 2a, the distance between the two vertices

√( (x+4)^2 + (y-6)^2 ) - √( ( x-4)^2 + y^2) = ±2a

but (2,0) lies on it, so
√( 6^2 + (-6)^2) - √( (-4)^2 + 0) = ±2a
√72 - 4 = ±2a
6√2 - 4 = 2a or 2a = 4-6√2

One wing:
√( (x+4)^2 + (y-6)^2 ) = √( ( x-4)^2 + y^2 ) + (6√2 - 4)
the other wing:
√( (x+4)^2 + (y-6)^2 ) = √( ( x-4)^2 + y^2 ) + (4 - 6√2 )

checking sofar with Wolfram:
http://www.wolframalpha.com/input/?i=plot+%E2%88%9A(+(x%2B4)%5E2+%2B+(y-6)%5E2+)+%3D++%E2%88%9A(+(+x-4)%5E2+%2B+y%5E2+)+%2B+(6%E2%88%9A2+-+4)
looks like a possible focus at (4,0)

http://www.wolframalpha.com/input/?i=plot+%E2%88%9A(+(x%2B4)%5E2+%2B+(y-6)%5E2+)+%3D++%E2%88%9A(+(+x-4)%5E2+%2B+y%5E2+)+%2B+(4+-+6%E2%88%9A2+)
possible focus at (-4,6)

square both sides:
x^2 + 8x+16 + y^2-12y+36 = x^2-8x+16 + (12√2-8)√((x-4)^2+y^2) + 72 - 48√2 + 16

16x - 12y + 36 = 4(3√2-2)√( (x-4)^2 + y^2)
phewww!!
4x - 3y + 9 = (3√2 - 2)√((x-4)^2 + y^2)

square both sides again.

16x^2 - 24xy + 72 + 9y^2 - 54y + 81 =
-12√2 x^2 + 22x^2 + 96√2 x - 176x - 12√2 y^2 + 22y^2 - 192√2 + 352

arhhhhggg!!

I think I made an error somewhere, there has to be an easier way.

Since this is a rotated hyperbola, have you studied the rotation matrix for conics?

Try this:
the centre must be the midpoint of AB which is (0,3)
take the length of AB, divide by 2, that will give you c
from above we found 2a = ....
thus we know "a"
using the property that a^2 + b^2 = c^2, you can find b

so the hyperbola in standard position would be
x^2 / a^2 - (y-3)^2 / b^2 = 1
plug in the values of "a" and "b"
we also know the slope of AB = 6/-8 = -3/4
so tanØ = -3/4
and sinØ = -3/5
and cosØ = 4/5

If I recall, the rotation matrix is
cosØ -sinØ
sinØ cosØ

There has to be examples in your text or your notes.
The last time I did this was about 45 years ago.