
Conic
Sections 


Hyperbola
and Line

Construction of tangents from a point outside the hyperbola

Properties of the hyperbola

The area of a triangle which the tangent at a point on the hyperbola forms with
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 area of a triangle which the tangent at a point on the hyperbola forms with
asymptotes, is of constant

value A
= a ·
b. Vertices of the triangle are the origin
O

and the intersections
A and
B of the tangent to the
hyperbola at
the point P_{0}(x_{0}, y_{0})
with
asymptotes. Then,

(1) t_{1 }::
b^{2}x_{0}x

a^{2}y_{0}y
= a^{2}b^{2}
the tangent

(2) y =
±
(b/a) · x
asymptotes

the solution of the system of the equations
(1)
and (2)
gives intersections A and
B. Plugging
(2) into
(1)






Since one vertex of the triangle is the origin
O(0, 0) then
the formula for the area, A_{D}=
(x_{2}y_{1}

x_{1}y_{2})/2
or 


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,
3/2)
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)]
= 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. 








Intermediate
algebra contents 



Copyright
© 2004  2020, Nabla Ltd. All rights reserved. 