
Conic
Sections 


Hyperbola
and Line

The equation of the tangent at the point on the hyperbola

Polar and pole of the hyperbola

Construction of the tangent at the point on the hyperbola

Hyperbola and line examples






The equation of the tangent at the point on the hyperbola

As we already mentioned, the points of contact of a line and the hyperbola can be obtained from the
corresponding formula for the ellipse by changing b^{2}
with b^{2}
thus 


the
tangency point or the point of contact. 

So, the intercept and slope of the
tangent 


or 
b^{2}x_{1}x

a^{2}y_{1}y
= a^{2}b^{2} 
the equation of the tangent at a point
P_{1}(x_{1},
y_{1})
on the hyperbola.



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'.





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 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°. 








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