
Conic
Sections 


Ellipse
and Line

Intersection of ellipse and line  tangency condition

Equation of the tangent at a point on the ellipse

Construction of the tangent at a point on the ellipse

Angle between the focal radii at a point of the ellipse

Ellipse and line examples







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 
(2) b^{2}x^{2} + a^{2}y^{2}
= a^{2}b^{2} 

by plugging (1)
into (2)
=>
b^{2}x^{2}
+ a^{2}(mx + c)^{2} = a^{2}b^{2} 
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, 
D = 0,
or 
a^{2}m^{2}
+ b^{2} = c^{2} 
a line is the tangent of the
ellipse and it is tangency
condition. 

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

D < 0, a line and an ellipse
do not intersect. 

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 the equation of the line through two given points determine the lines of the focal radii

r_{1} =
F_{1}P_{1}
and r_{2} =
F_{2}P_{1}, 

and the equation of the tangent at the point
P_{1},





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}
of 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 

If on the same way we calculate the interior
angle subtended by the focal radii at P_{1},
and which is the supplementary angle of the angle j, 

then compare with the
result which will we obtain by using the doubleangle formula for the angles
j_{1
}and j_{2}, 

To compare obtained results, we multiply both the
numerator and the denominator of the result for
the supplementary angle by
b^{2}, 

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

Ellipse and line examples

Example:
At a point A(c,
y
> 0) where c
denotes the focal distance,
on the ellipse
16x^{2}
+ 25y^{2} = 1600
drawn is a tangent to the ellipse, find the area of the triangle that tangent forms by the coordinate axes. 
Solution:
Rewrite the
equation of the ellipse to the standard form 16x^{2}
+ 25y^{2} = 1600  ¸
1600 

We calculate the ordinate of the point
A
by plugging the abscissa into equation of the ellipse

x =
6
=>
16x^{2}
+ 25y^{2} = 1600, 

or, as we know
that the point with the abscise
x = 
c
has the ordinate equal to the value of the
semilatus rectum,






The
area of the triangle formed by the tangent and the coordinate axes, 


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. 
Solution:
The tangency
points of tangents to the ellipse which are parallel with the given
line are, the
closest and the farthest points from the line.

Rewrite
the equation of the ellipse to determine its axes, 

Tangents and given line have the same slope, so 

Using the tangency condition, determine the intercepts
c, 

therefore, the
equations of tangents,





Solutions of the system of equations of tangents to the ellipse determine the points of contact, i.e., the
closest and the farthest point of the ellipse from the given line, thus 









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