|
|
|
|
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
b2
|
|
that is, a2(1
-
e2)
= -a2(e2
-
1) =
-b2, |
| 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 |
| b2x2
-
a2y2
= a2b2
|
| |
|
then if, a2m2
-
b2
> c2
the line intersects the hyperbola at two points,
|
|
a2m2
-
b2 = c2
the line is the tangent of the hyperbola,
|
|
a2m2
-
b2
< c2
the line and the hyperbola do not intersect. |
|
| Polar and pole of the hyperbola
|
If from a point
A(x0, y0), 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, |
| |
b2x0x
-
a2y0y
= a2b2 |
the equation of the
polar of the point A(x0, y0). |
|
|
| Construction of the tangent at the point on the hyperbola
|
| The tangent at the point
P1(x1,
y1)
on the hyperbola is the bisector of the angle F1P1F2
subtended by focal |
|
radii,
r1
and
r2 at
P1 .
|
|
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
F2,
and from F1as center, draw an arc of radius
2a.
|
|
These arcs intersect at points
S1
and
S2.
|
|
Tangents are the
perpendicular bisectors of the line segments F2S1
and
F2S2.
|
|
Tangents can also be drawn as lines through
A and
the intersection points of lines through F1S1
and F1S2,
with the hyperbola.
|
|
These intersections are at the same
time the points of contact
D1
and
D2.
|
|
 |
|
|
| 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, |
|
| 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. |
|
Solution:
Axes of the hyperbola we read from the standard form of equation,
|
|
4x2
-
9y2 = 36
| ¸36
|
|
|
We find tangents by solving the system of
equations,
|
|
(1) y =
mx + c
<=
A(0,
-3/2)
|
|
(2) a2m2
-
b2 = c2
<=
(1) c
= -3/2
|
|
|
|
9m2
-
4 = (-3/2)2,
9m2 =
25/4,
m1,2 =
±5/6
|
|
 |
|
| 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, |
 |
| Therefore,
the intersections S1(1,
-3/2)
and
S2(9,
6). |
| Then,
the area of the triangle AD=
(x2y1
-
x1y2)
/ 2
gives AD=
[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. |
|
|
|
|
|
|
|
|
|
|
| Pre-calculus contents
H |
|
 |
|
| Copyright
© 2004 - 2020, Nabla Ltd. All rights reserved. |
