
Conic
Sections 


Hyperbola
and Line

Construction of tangents from a point outside the hyperbola

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






Construction of tangents from a point outside the hyperbola 
With A as center draw an arc through
F_{2},
and from F_{1}as center, draw an arc of radius
2a.

These arcs intersect at points
S_{1}
and
S_{2}.

Tangents are the
perpendicular bisectors of the line segments F_{2}S_{1}
and
F_{2}S_{2}.

Tangents can also be drawn as lines through
A and
the intersection points of lines through F_{1}S_{1}
and
F_{1}S_{2},
with the hyperbola.

These intersections are at the same
time the points of contact
D_{1}
and
D_{2}.





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
P_{0}, we can construct the tangent by drawing the
parallel to the asymptote
y = (b/a) · x
through P_{0}
to D. Mark endpoint
B
of segment OB
taking D
as the midpoint. Thus, the line segment P_{0}B 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,
DP_{0}C,
DBP_{0} and
CP_{0}A,
are congruent, it follows that the area of the parallelogram
ODP_{0}C
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
= a^{2}
/ 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 semiaxis
a
such that the line
5x

4y

16 = 0 be the tangent of the hyperbola 
9x^{2}

a^{2}y^{2} = 9a^{2}. 
Solution:
Rewrite the equation
9x^{2}

a^{2}y^{2} = 9a^{2}  ¸
9a^{2}


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) c_{H} =
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
3x^{2}

4y^{2} = 12 which is parallel to the line
x +
y = 0.

Solution:
Rewrite the equation of the hyperbola

3x^{2}

4y^{2} = 12
 ¸12


The slope of the normal is equal to the slope of
the given line,

y =
x
=>
m
= 1,
m_{t} =
1/m_{n},
so m_{t} =
1

applying the tangency condition

a^{2}m^{2}

b^{2} = c^{2}
<= m_{t} =
1,
a^{2} =
4 and
b^{2} = 3

4·(1)^{2}

3 = c^{2}
=> c_{1,2} = ±1

tangents, t_{1
}::
y =
x
+ 1 and
t_{2
}::
y =
x

1.

The points of
tangency,





The
equations of the normals, 
D_{1}(4,
3)
and m =
1
=>
y 
y_{1} = m ·(x
x_{1}),
y +
3 = 1·(x
 4)
or n_{1}_{
}::
y = x 
7, 
D_{2}(4,
3)
and m =
1 =>
y 
y_{1} = m ·(x
x_{1}),
y 
3 = 1·(x
+ 4)
or n_{2}_{
}::
y =
x + 7. 

Example:
From the point A(0,
3/2)
drawn are tangents to the hyperbola 4x^{2}

9y^{2} = 36, find the
equations of the tangents and the area of the triangle which both tangents form with asymptotes. 
Solution:
Axes of the hyperbola we read from the standard form of equation,

4x^{2}

9y^{2} = 36
 ¸36


We find tangents by solving the system of
equations,

(1) y =
mx + c
<=
A(0,
3/2)

(2) a^{2}m^{2}

b^{2} = c^{2}
<=
(1) c
= 3/2


9m^{2}

4 = (3/2)^{2},
9m^{2} =
25/4,
m_{1,2} =
±5/6




thus,
the equations of the tangents, 
t_{1}_{
}::
y =
5/6x

3/2 or 5x

6y 
9 = 0 and
t_{2 }_{
}::
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, 
A_{D}=
(x_{2}y_{1}

x_{1}y_{2})/2,
where S_{1}(x_{1},
y_{1}) and S_{2}(x_{2},
y_{2})
are the intersections (third vertex is the origin (0, 0)). 
By solving system of
equations, 

Therefore,
the intersections S_{1}(1,
2/3)
and
S_{2}(9,
6). 
Then,
the area of the triangle A_{D}=
(x_{2}y_{1}

x_{1}y_{2})
/ 2
gives A_{D}=
[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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