Section 10-4
Hyperbolas
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A hyperbola is the set of all points in the plane in which the difference of the distances
from two fixed points called foci is constant. The center of the hyperbola is the
midpoint of the line segment whose endpoints are the foci. The point on each branch
of the hyperbola that is nearest the center is called a vertex. The asymptotes of the
hyperbola are lines that the curve approaches as it recedes from the center. There are
two axes of symmetry. The line segment connecting the vertices is called the
transverse axis and has a length of 2a units. The segment perpendicular to the
transverse axis through the center is called the conjugate axis and has length 2b units.
PF1-PF2
=
QF1-QF2
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The standard form of the hyperbola is similar
to the ellipse form except it has subtraction
instead of addition.
b g 
x h
a2
2
by  k g
b2
2
1
2


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by  k g  bx  h g
b
a
2
2
and
2
2
1
2
The General Form: Ax + Bxy + Cy + Dx + Ey + F = 0
Quick way to differentiate the equations for the two shapes: Remember
for a hyperbola A and C have different signs. Whereas for an ellipse A
an d C h ave th e sam e sign an d A ≠C .
Also notice the difference in a, b, and c.
2
2
2
2
2
2
For an ellipse c = a – b . For a hyperbola c = a + b
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Find the equation of the hyperbola with foci (1, -5) and (1,1) and
whose transverse axis is 4 units long( this is 2a).
On the foci it is the y that has changed. So h= 1 and we know we
are working with the equation
by  k g 
a
2


bx  hg
2
1
and since the transverse axis is
parallel to the y axis (x=1), we know that a=2.
Now we know k+c=1 and k-c =-5. solving this system of
2
equations, k=-2 and C=3. We can find b by 9= 4+b b= 5
2
by  2 g
4
b2

b x  1g
5
2
1
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Find the coordinates of the center, foci, and vertices, and the equations of
the asymptotes of the graph
Then graph the equation.
b
x 5
25
g  b y  1g
2
9
2
1
From the equation we know the center (5,-1), the foci (h±c, -1), the vertices
(h±a, k), and the asymptotes y+1=± b/a (x-5) .
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We know a = 5 and b = 3 so c = 34 . The foci are (5± 34 , -1). The
vertices are (5±5, -1) = (0,-1) and (10, -1). The asymptotes are
y+1=± 3 (x-5)
up 3
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To draw the graph, first make your box

5
5 left
x (5,-1)
5 right
down 3
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Next draw the asymptotes
Then draw the hyperbolas
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Find the coordinates of the center, foci, and
vertices, and the equations of the asymptotes of
2
2
the graph of 4x –y +24x +4y + 28 = 0
Step 1: Complete the square to put it in standard form
2
2
4x +24x – (y -4y) = -28
4 ( x 2 +6x) – (y-2) 2 =-28 - 4
2
4 ( x+3) - (y – 2) 2 = -28 - 4 + 36 = 4
2
2
(x+3) / 1 – (y – 2) /4 = 1
2
Center (-3,2) Foci
a + b 2 = c 2 1+4 = 5
So foci (-3± 5 . 2)
Vertices (-2,2) (-4, 2) (h±a, k)
Equations of asymptotes y-2 = ±2(x+3)
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In the standard form of the equation of a
hyperbola, if a=b, the graph is an equilateral
hyperbola. The slopes of the equations of the
two asymptotes are negative reciprocals 1, -1
and they are perpendicular.
Question: When drawing the box, what type of
shape is it?
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A special case called the rectangular hyperbola
is where the coordinate axes ( x and y ) are the
asymptotes. The general equation of the
rectangular hyperbola is xy = c where c is a
constant. In General form it means that
A=C=D=E=0 and we only have B and F.
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Graph xy = 36. Since c is positive, the
hyperbola lies in the first and third quadrants.
The vertices must satisfy the equation and also
their graph must be points on the traverse axis
which is along the graph of y= x. So the
vertices are (6,6) and (-6,-6)
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6
*
*
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Like an ellipse. The shape of a hyperbola is
determined by its eccentricity which is again
defined as е = c/a.