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| Conic
Sections |
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Hyperbola
and Line
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Construction of tangents from a point outside the hyperbola
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Properties of the hyperbola
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The
parallels to the asymptotes through the tangency point intersect
asymptotes |
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The equation of the equilateral or rectangular hyperbola with the
coordinate axes as its
asymptotes |
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Hyperbola and line examples
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| Construction of tangents from a point outside the hyperbola |
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With A as center draw an arc through
F2,
and from F1as center, draw an arc of radius
2a.
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These arcs intersect at points
S1
and
S2.
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Tangents are the
perpendicular bisectors of the line segments F2S1
and
F2S2.
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Tangents can also be drawn as lines through
A and
the intersection points of lines through F1S1
and
F1S2,
with the hyperbola.
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These intersections are at the same
time the points of contact
D1
and
D2.
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| Properties of the hyperbola
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| -
The tangency point bisects the line
segment AB
of
the tangent between
asymptotes. |
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The abscissa of the midpoint of the segment
AB,
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| equals the abscissa of the tangency point.
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-
The parallels to the asymptotes through the
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tangency point intersect asymptotes at the points,
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C
and
D such
that,
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OC
= AC and
OD
= BD
.
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| Therefore, if given are asymptotes and the tangency point
P0, we can construct the tangent by drawing the
parallel to the asymptote
y = (b/a) · x
through P0
to D. Mark endpoint
B
of segment OB
taking D
as the midpoint. Thus, the line segment P0B determines the tangent line. |
| On a similar way we could determine intersection
A,
of the tangent and another asymptote, using point C. |
| Since
triangles, ODC,
DP0C,
DBP0 and
CP0A,
are congruent, it follows that the area of the parallelogram
ODP0C
is equal to half of the area of the triangle
OBA, i.e.,
A
= (a ·
b) / 2. |
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| Using this property we can derive
the equation of the equilateral or rectangular hyperbola with the
coordinate |
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axes as its
asymptotes.
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As the asymptotes of an equilateral hyperbola are
mutually perpendicular then the given parallelogram is
the rectangle.
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| And since the axes of the equilateral hyperbola are equal
that is a
= b, then
the area A
= a2
/ 2.
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Then, for every point in the new coordinate
system
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If we now change the coordinates into
x
and y,
and denote the constant by c, obtained is
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the equation of
the equilateral or rectangular hyperbola
with the coordinate axes as its
asymptotes.
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| Hyperbola and line examples
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| Example:
Determine the semi-axis
a
such that the line
5x
-
4y
-
16 = 0 be the tangent of the hyperbola |
| 9x2
-
a2y2 = 9a2. |
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Solution:
Rewrite the equation
9x2
-
a2y2 = 9a2 | ¸
9a2
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| and the equation of the tangent
5x
-
4y
-
16 = 0
or |
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| Then,
plug the slope and the intercept into tangency condition,
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| Therefore,
the given line is the tangent of the hyperbola |
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| Example:
The line 13x
-
15y
-
25 = 0 is the tangent of a hyperbola with linear eccentricity (half the focal
distance) cH =
Ö41.
Write the equation of the hyperbola.
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| Solution:
Rewrite the equation 13x
-
15y
-
25 = 0
or |
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| Using
the linear eccentricity |
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| and
the tangency condition |
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| Thus,
the equation of the hyperbola, |
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| Example:
Find the normal to the hyperbola
3x2
-
4y2 = 12 which is parallel to the line
-x +
y = 0.
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Solution:
Rewrite the equation of the hyperbola
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3x2
-
4y2 = 12
| ¸12
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The slope of the normal is equal to the slope of
the given line,
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y =
x
=>
m
= 1,
mt =
-1/mn,
so mt =
-1
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applying the tangency condition
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a2m2
-
b2 = c2
<= mt =
-1,
a2 =
4 and
b2 = 3
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4·(-1)2
-
3 = c2
=> c1,2 = ±1
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tangents, t1
::
y =
-x
+ 1 and
t2
::
y =
-x
-
1.
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The points of
tangency,
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| The
equations of the normals, |
| D1(4,
-3)
and m =
1
=>
y -
y1 = m ·(x
-x1),
y +
3 = 1·(x
- 4)
or n1
::
y = x -
7, |
| D2(-4,
3)
and m =
1 =>
y -
y1 = m ·(x
-x1),
y -
3 = 1·(x
+ 4)
or n2
::
y =
x + 7. |
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| Example:
From the point A(0,
-3/2)
drawn are tangents to the hyperbola 4x2
-
9y2 = 36, find the
equations of the tangents and the area of the triangle which both tangents form with asymptotes. |
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Solution:
Axes of the hyperbola we read from the standard form of equation,
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4x2
-
9y2 = 36
| ¸36
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We find tangents by solving the system of
equations,
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(1) y =
mx + c
<=
A(0,
-3/2)
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(2) a2m2
-
b2 = c2
<=
(1) c
= -3/2
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9m2
-
4 = (-3/2)2,
9m2 =
25/4,
m1,2 =
±5/6
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| thus,
the equations of the tangents, |
| t1
::
y =
5/6x
-
3/2 or 5x
-
6y -
9 = 0 and
t2
::
y =
-5/6x
-
3/2 or 5x
+ 6y + 9 = 0. |
| The area of the triangle that tangents form with asymptotes we calculate
using the formula, |
| AD=
(x2y1
-
x1y2)/2,
where S1(x1,
y1) and S2(x2,
y2)
are the intersections (third vertex is the origin (0, 0)). |
| By solving system of
equations, |
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| Therefore,
the intersections S1(1,
-2/3)
and
S2(9,
6). |
| Then,
the area of the triangle AD=
(x2y1
-
x1y2)
/ 2
gives AD=
[1·6
-
9·(-2/3)]
/ 2
= 6 square units. |
We can
get the same result using the property that the area of the triangle which the tangent form
with asymptotes of the hyperbola is of the constant value A
= a ·
b, so that A =
3 ·
2
= 6. |
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| Conic
sections contents |
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