A hyperbola with one of the vertex at (5,6), the centre
at (0,6) and one of the slope is equal to 2.
a) Sketch the graph.
b) Find the equation.
c) Locate the foci.

a) Sketch the graph. y=2x
(0,16)
B1
(­5√5, 6)
F2 A A1 F1 (5√5, 6)
a = 5 2
C (5,6)
(­5,6) (0,6)
b = 10
c = 5√5
B2
(0,­4)

a is the semi­transverse axis. It is the
distance from the centre to one of the
vertices.
b is the semi­conjugate axis. It is
a = 5
the distance from the centre to
b = 10
one of the points B1 or B2.
c = 5√5
c is the distance between the centre
and one of the foci. It is also equal to
the distance between the vertex to one
of the points B1 or B2.

We found the distance of the semi­conjugate by
multiplying the y­coordinate of the vertex to the rise of
the slope.
b = 10
We also figured out the distance of the foci by using
the pythagorean theorem.
c2 = a2 + b2

b) Find the equation.
We know from the graph we made
centre (h,k)
that the we had a horizontal hyperbola.
C (0,6) Therefore the standard equation would
be:
a = 5
2 2
(x­h) _ (y­k) = 1
b = 10
a2 b2
c = 5√5

By substituting the values the resulting equation
would be:
2 2
(x) _ (y­6) = 1
25 100

We already figured out where are the locations of the
foci when we graphed it.
Given that we know the semi­transverse axis
(a) and semi­conjugate axis (b), Use the
pythagorean theorem to find (c).
2
c = 125
2 2 2
c = a + b
2 2 2
c = 5 + 10 c = √125 or 5√5
2
c = 25 + 100

The foci's locations are:
F2
F1
(­5√5, 6)
(5√5, 6)