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Analytic geometry
- Conic sections
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Hyperbola
and line
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Hyperbola and line relationships
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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
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|m|
<
b/a but
if |m|
> b/a
then,
the line
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y =
mx does not intersect the hyperbola at all.
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The diameters of a hyperbola are straight lines
passing through its center.
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The asymptotes divide these two pencils of diameters
into, one which intersects the curve at two points, and
the
other which do not intersect.
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A diameter of a conic section is a line which passes
through the midpoints of parallel chords.
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Conjugate diameters of the hyperbola (or the ellipse) are two diameters such that each bisects all chords
drawn parallel to the other.
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As the equation of a hyperbola can be obtained from the equation of an ellipse by changing the sign of
b2
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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, to examine conditions which determine position of a line in relation to a hyperbola
we
solve the system of
equations, |
y =
mx + c
then if,
a2m2
-
b2
> c2
the line intersects the hyperbola at two points,
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b2x2
-
a2y2
= a2b2
a2m2
-
b2
< c2
the line and the hyperbola do not intersect. |
a2m2
-
b2 = c2
the line is the tangent of the hyperbola,
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Condition for a line to be the tangent to the hyperbola -
tangency condition
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A
line is the tangent to the hyperbola if
a2m2
-
b2 = c2,
this equality is called the tangency
condition. |
Regarding
the asymptotes, to which c
= 0, this condition gives
|m|
= b/a
and that is why we can say that the hyperbola touches the asymptotes at infinity. |
The tangency condition also
states that the slopes of the tangents
will satisfy the condition if |
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That is, the tangents to the hyperbola can only be parallel to a line belonging to the pencil of lines that do not
intersects the hyperbola. |
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The equation of the tangent at the point on the hyperbola
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The points of contact of a line and the hyperbola can be obtained from the corresponding formula for the ellipse by changing
b2
with -b2
thus, |
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the
tangency point or the point of contact. |
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Thus,
the intercept and slope of the
tangent |
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or b2x1x
-
a2y1y
= a2b2
the equation of the tangent at a point P1(x1,
y1)
on the hyperbola.
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Polar and pole of the hyperbola
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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). |
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Construction of the tangent at the point on the hyperbola
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The tangent at the point
P1(x1,
y1)
on the hyperbola is the bisector of the angle F1P1F2
subtended by |
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Construction of tangents from a point outside the hyperbola
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With A as center draw an arc through
F2,
and from F1as center, draw an arc of radius
2a.
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These arcs intersect at points
S1
and
S2.
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Tangents are the
perpendicular bisectors of the line segments F2S1
and
F2S2.
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Tangents can also be drawn as lines through
A and
the intersection points of lines through F1S1
and F1S2,
with the hyperbola.
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These intersections are at the same
time the points of contact
D1
and
D2.
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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.
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Solution:
Rewrite the equation 13x
-
15y
-
25 = 0
or |
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Using
the linear eccentricity |
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and
the tangency condition |
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hus,
the equation of the hyperbola, |
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Example:
Find the normal to the hyperbola
3x2
-
4y2 = 12 which is parallel to the line
-
x +
y = 0.
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Solution: Rewrite the equation of the hyperbola
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3x2
-
4y2 = 12
| ¸12
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The slope of the normal is equal to the slope of
the given line,
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y =
x
=> m
= 1,
mt =
-1/mn,
so mt =
-1
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applying the tangency condition
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a2m2
-
b2 = c2
<= mt =
-1,
a2 =
4,
b2 = 3
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4·(-1)2
-
3 = c2
=> c1,2 = ±1
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tangents
t1
::
y =
-x
+ 1,
t2
::
y =
-x
-
1.
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The points of
tangency,
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The
equations of the normals,
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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. |
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Contents
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