Find an equation of the hyperbola such taht for any point on the hyperbola.

Question

the difference between its distance from the points (2,2) and (10,2) is 6.

Answer 1

This hyperbola has a horizontal transverse axis, i'e', it
looks sort of like this: ")(", so it has equation of the form

(x - h)²   (y - k)²
———————— - ———————— = 1
   a²         b²

where (h-a,k) and (h+a.k) are the left and right vertices,
respectively

where (h-c,k) and (h+c,k) are the left and right foci,
respectively

where c = absolute value distance between the foci and the
center.  

where a² + b² = c²

where a = absolute value distance between the center and the
vertex = semi-transverse axis.

where 2a = transverse axis

where b = semi-conjugate axis

and where 2b = conjugate axis

The given points are the foci.  The center (h,k) of the
hyberbola is the midpoint between these foci or 
(h,k) = ((2+10)/2, (2+2)/2)) = (6,2).
So far we have

(x - 6)²   (y - 2)²
———————— - ———————— = 1
   a²         b²

The vertices are " a " units from the center, so their
coordinates are

(h-a, 2) and (h+a, 2)

and since h = 6, the vertices are

(6-a, 2) and (6+a, 2)

Also c = absolute value distance between center and focus,
and the distance between (6,2) and (2,2) is 4, so c=4

The vertices are themselves points on the hyperbola.
Therefore the difference (in absolute value) between each one
of these vertices' distance from the points (2,2) and (10,2)
is 6.

The vertices are between the foci, so

The distance between (6-a, 2) and (2,2) is (6-a)-2 or 4-a

The distance between (6-a, 2) and (10,2) is 10-(6-a) or a+4

The difference between these in absolute value = (a+4)-(4-a)
or 2a

So 2a = 6 or a = 3 

Therfore we have

(x - 6)²   (y - 2)²
———————— - ———————— = 1
   3²         b²

or

(x - 6)²   (y - 2)²
———————— - ———————— = 1
   9          b²

All that's left is b², and we get that from

a² + b² = c²
3² + b² = 4²
 9 + b² = 16
     b² = 7

So we replace b² by 7 and we are done:

 (x - 6)²   (y - 2)²
———————— - ———————— = 1
   9           7