
Conic
Sections 


Hyperbola
and Line

Hyperbola and line relationships

Hyperbola and line examples






Hyperbola and line examples

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. 

Example:
Find the angle
between the ellipse, which passes through points, A(Ö5,
4/3) and B(1,
4Ö2/3), and
the hyperbola whose asymptotes are y = ±
x/ 2
and the linear eccentricity or half of the focal distance c
= Ö5. 
Solution:
Find
the equation of the ellipse by solving the system of equations, 


thus,
the equation of the ellipse 


The equation of the hyperbola by solving the system of
equations,





Therefore, equation of the hyperbola 


Angle between curves is the angle between tangents at intersection of the curves. By solving the system of
equations of the curves we obtain the points of intersection, 
(1)
4x^{2}
+ 9y^{2} = 36 (2)
=> (1)
4(4y^{2}
+ 4) +
9y^{2} = 36,
25y^{2} = 20,
y_{1,2} = ±2/Ö5,
x_{1,2} = ±6/Ö5 
(2) x^{2}

4y^{2} = 4
=>
x^{2} = 4y^{2}
+ 4 
Tangent
of the ellipse and the hyperbola at the intersection S_{1}(6/Ö5,
2/Ö5), 
S_{1}(6/Ö5,
2/Ö5)
=> t_{e
}::
b^{2}x_{1}x
+ a^{2}y_{1}y
= a^{2}b^{2},
4x + 3y
= 6/Ö5
or y =
(4/3)x
+ 2Ö5, 
S_{1}(6/Ö5,
2/Ö5)
=>
t_{h}_{
}::
b^{2}x_{1}x

a^{2}y_{1}y
= a^{2}b^{2},
3x 
4y = 2/Ö5
or y
= (3/4)x 
Ö5/2. 

fulfilled is the perpendicularity condition.
Therefore, the angle between curves j = 90°. 








College
algebra contents E 



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