

Analytic geometry
 Conic sections


::
Intersection of ellipse and line  tangency condition

Common points of a line and an ellipse
we find by solving their equations as a system of two equations in two
unknowns, x
and y, 
(1) y = mx + c
by plugging (1)
into (2)
=>
b^{2}x^{2}
+ a^{2}(mx + c)^{2} = a^{2}b^{2} 
(2) b^{2}x^{2} + a^{2}y^{2}
= a^{2}b^{2}
and^{
}after
rearranging (a^{2}m^{2}
+ b^{2})·x^{2} + 2a^{2}mc·x
+ a^{2}c^{2} 
a^{2}b^{2} = 0 
obtained
is the quadratic equation in x.
Thus, the coordinate of intersections are, 

Using the above solutions follows that a line and an ellipse can have one of three possible mutual positions
depending of the value of the discriminant
D
= a^{2}m^{2}
+ b^{2} 
c^{2}. Thus, if 
D > 0,
a line and an ellipse
intersect, and if D < 0,
a line and an ellipse
do not intersect, 
while
if
D = 0,
or a^{2}m^{2}
+ b^{2} = c^{2}
a line is the tangent to the
ellipse thus, it is the tangency
condition.

The line touches the ellipse at the tangency point whose coordinates are: 


Equation of the tangent at a point on the ellipse

In the equation
of the line y

y_{1} = m(x

x_{1})
through a given point P_{1}, the slope m
can be determined using known coordinates (x_{1},
y_{1}) of the point of tangency, so


b^{2}x_{1}x
+ a^{2}y_{1}y
= b^{2}x_{1}^{2}
+ a^{2}y_{1}^{2},
since b^{2}x_{1}^{2}
+ a^{2}y_{1}^{2} =
a^{2}b^{2}
is the condition that P_{1}
lies on the ellipse

then
b^{2}x_{1}x
+ a^{2}y_{1}y
= a^{2}b^{2
}is the
equation of the tangent at the point P_{1}(x_{1},
y_{1})
on the ellipse.


Construction of the tangent at a point on the ellipse

Draw a circle of a radius
a concentric to the ellipse. Extend the ordinate of the given point to find

intersection with the circle. The tangent of the circle at
P_{c}
intersects the xaxis at
P_{x}. The tangent to the ellipse at the point P_{1
}on the
ellipse intersects the xaxis at the same point.

To prove this, find the
xintercept of each tangent
analytically.

Therefore, in both equations of tangents set y = 0 and
solve for x,


it is the xintercept of the tangent
t_{c}
and the tangent t_{e}.





Angle between the focal radii at a point of the ellipse

Let
prove that the tangent at a point
P_{1
}of the ellipse is perpendicular to the bisector of the angle between the focal radii
r_{1
}and r_{2}.
Coordinates of points,
F_{1}(c,
0),
F_{2}(c,
0) and P_{1}(x_{1},
y_{1}) plugged
into


By plugging the slopes of these tree lines into the formula for calculating the angle between lines we find the
exterior angles
j_{1
}and j_{2}
subtended by these lines at P_{1}. 
Thus, using the condition
b^{2}x_{1}^{2}
+ a^{2}y_{1}^{2}
= a^{2}b^{2}, that the point lies on the ellipse, obtained is 

Therefore, the normal at the point
P_{1}
of the ellipse bisects the interior angle between its focal radii.


Example:
Find a point on the ellipse
x^{2}
+ 5y^{2} = 36 which is the closest, and which is the farthest from the
line 6x + 5y  25 =
0.


Tangents to an ellipse from a
point outside the ellipse  use of the tangency condition

Coordinates of the point
A(x, y), from which we draw tangents to an ellipse, must satisfy
equations of the tangents, y
= mx + c and their slopes and intercepts,
m and
c, must satisfy the
condition of tangency therefore, using the system of equations, 
(1)
y = mx + c
<=
A(x, y) 
(2)
a^{2}m^{2} + b^{2}
= c^{2}
determined are equations of the tangents from a point A(x, y)
outside the ellipse.


Construction of tangents from a point outside the ellipse

With
A
as center, draw an arc through F_{2}, and from
F_{1}
as center, draw an arc of the radius 2a. Tangents are
then the perpendicular bisectors of the line segments,
F_{2}S_{1} and
F_{2}S_{2}.

We can also draw tangents as lines through A
and the intersection points of the segments F_{1}S_{1} and
F_{1}S_{2}
and the ellipse. 
Thus, these intersections are the tangency points
of the tangents to the ellipse. 
Explanation of the construction lies at the fact that, 
F_{1}S_{1} =
F_{1}D_{1} +
D_{1}S_{1} =
F_{1}D_{1} +
D_{1}F_{2} = 2a

according to the definition of the ellipse, as well as
the point A
is equidistant from points F_{2} and
S_{1},
since the point S_{1}
lies on the arc drawn from A
through F_{2}.





Example:
Determine equation of the ellipse which the line
3x +
10y = 25 touches at the point
P(3,
8/5).

Solution: As the given line is the tangent to the ellipse,
parameters, m
and c
of the line must satisfy the tangency condition, and the point P
must satisfy the equations of the line and the ellipse, thus 


Example:
Find the equations of the common tangents of the curves
4x^{2}
+ 9y^{2} = 36 and
x^{2}
+ y^{2} = 5.

Solution: The common tangents of the ellipse and the circle
must satisfy the tangency conditions of these curves, thus




Polar and pole of the ellipse

If from a point A(x_{0}, y_{0}), exterior to the ellipse, drawn are tangents, then the secant line passing through the
contact points, D_{1}(x_{1}, y_{1}) and
D_{2}(x_{2}, y_{2}) is the polar of the point
A. The point
A is called the
pole of 
the
polar,
as shows the right figure. 
Coordinates of the point
A(x_{0}, y_{0}) must satisfy the equations of tangents, thus 
t_{1} ::
b^{2}x_{0}x_{1}
+ a^{2}y_{0}y_{1}
= a^{2}b^{2} 
t_{2} ::
b^{2}x_{0}x_{2}
+ a^{2}y_{0}y_{2}
= a^{2}b^{2} 
and after
subtracting
t_{1
}
t_{2} 
b^{2}x_{0}(x_{2}

x_{1})
+ a^{2}y_{0}(y_{2}

y_{1})
= 0

obtained
is 





the slope of the secant line through points of contact
D_{1}and
D_{2}.

Thus, the equation of the secant line


and after rearranging 

b^{2}x_{0}x
+ a^{2}y_{0}y
= b^{2}x_{0}x_{1}
+ a^{2}y_{0}y_{1},
since b^{2}x_{0}x_{1}
+ a^{2}y_{0}y_{1}
= a^{2}b^{2} 
follows
b^{2}x_{0}x
+ a^{2}y_{0}y
= a^{2}b^{2}
the equation of
the polar p
of the
point A(x_{0}, y_{0}). 

Example:
The line x
+ 14y

25 = 0 is the polar of the ellipse
x^{2}
+ 4y^{2} = 25.
Find coordinates of the pole.

Solution: Intersections of the polar and the ellipse are points
of contact of tangents drawn from the pole P
to ellipse, thus solutions of the system of equations, 
(1)
x
+ 14y

25 = 0 => x
= 25  14y
=> (2) 
(2)
x^{2}
+ 4y^{2} = 25

(25  14y)^{2}
+ 4y^{2} = 25

2y^{2}

7y + 6 = 0,
y_{1} =
3/2 and
y_{2} =
2

y_{1}
and y_{2}
=>
x
= 25  14y,
x_{1} =
4 and
x_{2} =
3.

Thus, the points of
contact D_{1}(4,
3/2) and
D_{2}(3,
2).

The
equations of the tangents at D_{1}
and
D_{2},





The solutions of the system of equations t_{1}
and t_{2
}are the coordinates of the
pole P(1, 7/2). 















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