Find an equation of the hyperbola such taht for any point on
the hyperbola. the difference between its distance from the
points (2,2) and (10,2) is 6.
This hyperbola has a horizontal transverse axis, i'e', it
looks sort of like this: ")(", so it has equation of the form
(x - h)ｲ (y - k)ｲ
覧覧覧覧 - 覧覧覧覧 = 1
aｲ bｲ
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 aｲ + bｲ = cｲ
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)ｲ (y - 2)ｲ
覧覧覧覧 - 覧覧覧覧 = 1
aｲ bｲ
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)ｲ (y - 2)ｲ
覧覧覧覧 - 覧覧覧覧 = 1
3ｲ bｲ
or
(x - 6)ｲ (y - 2)ｲ
覧覧覧覧 - 覧覧覧覧 = 1
9 bｲ
All that's left is bｲ, and we get that from
aｲ + bｲ = cｲ
3ｲ + bｲ = 4ｲ
9 + bｲ = 16
b! = 7
So we replace bｲ by 7 and we are done:
(x - 6)ｲ (y - 2)ｲ
覧覧覧覧 - 覧覧覧覧 = 1
9 7
Edwin
AnlytcPhil@aol.com
