|
|
Hyperbola
and Line
|
Properties of the hyperbola
|
The parallels to the asymptotes through the tangency point intersect asymptotes |
The equation of the equilateral or rectangular hyperbola with the
coordinate axes as its
asymptotes |
Hyperbola and line examples
|
|
|
|
|
|
Properties of the hyperbola
|
-
The tangency point bisects the line
segment AB
of
the tangent between
asymptotes. |
The abscissa of the midpoint of the segment
AB,
|
|
equals the abscissa of the tangency point.
|
-
The parallels to the asymptotes through the
tangency point intersect asymptotes at the points,
C
and
D such
that,
|
OC
= AC and
OD
= BD
.
|
|
|
|
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. |
|
Using this property we can derive
the equation of the equilateral or rectangular hyperbola with the
coordinate |
axes as its
asymptotes.
|
As the asymptotes of an equilateral hyperbola are
mutually perpendicular then the given parallelogram is
the rectangle.
|
And since the axes of the equilateral hyperbola are equal
that is a
= b, then
the area A
= a2
/ 2.
|
Then, for every point in the new coordinate system
|
|
If we now change the coordinates into
x
and y,
and denote the constant by c, obtained is
|
|
the equation of
the equilateral or rectangular hyperbola
with the coordinate axes as its
asymptotes.
|
|
|
|
|
Hyperbola and line examples
|
Example:
Determine the semi-axis
a
such that the line
5x
-
4y
-
16 = 0 be the tangent of the hyperbola |
9x2
-
a2y2 = 9a2. |
Solution:
Rewrite the equation
9x2
-
a2y2 = 9a2 | ¸
9a2
|
|
and the equation of the tangent
5x
-
4y
-
16 = 0
or |
|
|
Then,
plug the slope and the intercept into tangency condition,
|
|
Therefore,
the given line is the tangent of the hyperbola |
|
|
|
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.
|
Solution:
Rewrite the equation 13x
-
15y
-
25 = 0
or |
|
|
Using
the linear eccentricity |
|
|
and
the tangency condition |
|
Thus,
the equation of the hyperbola, |
|
|
|
Example:
Find the normal to the hyperbola
3x2
-
4y2 = 12 which is parallel to the line
-x +
y = 0.
|
Solution:
Rewrite the equation of the hyperbola
|
3x2
-
4y2 = 12
| ¸12
|
|
The slope of the normal is equal to the slope of
the given line,
|
y =
x
=>
m
= 1,
mt =
-1/mn,
so mt =
-1
|
applying the tangency condition
|
a2m2
-
b2 = c2
<= mt =
-1,
a2 =
4 and
b2 = 3
|
4·(-1)2
-
3 = c2
=> c1,2 = ±1
|
tangents, t1
::
y =
-x
+ 1 and
t2
::
y =
-x
-
1.
|
The points of
tangency,
|
|
|
|
|
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, |
|
|
|
|
|
|
|
|
Pre-calculus contents
H |
|
|
|
Copyright
© 2004 - 2020, Nabla Ltd. All rights reserved. |