

Hyperbola
and Line

Hyperbola and line relationships

Polar and pole of the hyperbola

Construction of the tangent at the point on the hyperbola

Construction of tangents from a point outside the hyperbola

Hyperbola and line examples






Hyperbola and line relationships

Let examine relationships between a hyperbola and a line passing through the center of the hyperbola, i.e., the
origin. A line y =
mx intersects the hyperbola at two
points if

the slope m
<
b/a but
if m
> b/a
then,

the line y =
mx does not intersect the hyperbola at all.

The diameters of a hyperbola are straight lines
passing through its center.

The asymptotes divide these two pencils of diameters
into, one which intersects the curve at two points, and
the
other which do not intersect.

A diameter of a conic section is a line which passes
through the midpoints of parallel chords.

Conjugate diameters of the hyperbola (or the ellipse)
are two diameters such that each bisects all chords drawn parallel to the other.




As the equation of a hyperbola can be obtained from the equation of an ellipse by changing the sign of
b^{2}

that is, a^{2}(1

e^{2})
= a^{2}(e^{2}

1) =
b^{2}, 
this way, we can use other formulas relating to the ellipse to obtain corresponding
formulas for the hyperbola. 
Therefore, when we examine conditions which determine position of a line in relation to a hyperbola that is, 
when
solve the system of
equations,
y =
mx + c 
b^{2}x^{2}

a^{2}y^{2}
= a^{2}b^{2}


then if, a^{2}m^{2}

b^{2}
> c^{2}
the line intersects the hyperbola at two points,

a^{2}m^{2}

b^{2} = c^{2}
the line is the tangent of the hyperbola,

a^{2}m^{2}

b^{2}
< c^{2}
the line and the hyperbola do not intersect. 

Polar and pole of the hyperbola

If from a point
A(x_{0}, y_{0}), exterior to the hyperbola, drawn are tangents, then the secant line passing through
the contact points, is the polar of the point A.
The point A
is called the pole of the polar. 
The equation of the
polar is derived the same way as for the ellipse, 

b^{2}x_{0}x

a^{2}y_{0}y
= a^{2}b^{2} 
the equation of the
polar of the point A(x_{0}, y_{0}). 


Construction of the tangent at the point on the hyperbola

The tangent at the point
P_{1}(x_{1},
y_{1})
on the hyperbola is the bisector of the angle F_{1}P_{1}F_{2}
subtended by focal 
radii,
r_{1}
and
r_{2} at
P_{1} .

The proof shown for the ellipse applied to
the hyperbola gives,


or 



See the title
'
The angle between the focal radii at a point of the
ellipse'.





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}.





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. 








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