We start with a circle C with radius a.
x2+ y2 = a2
If we multiply the y-value of all points P' on the circle by a fixed value b/a, then each
point P' is transformed in a new point P.
The set of all these points P is an ellipse.
P(x,y) is on the ellipse
<=>
P'(x, (a/b) y ) is on C
<=>
x2 + (a/b)2 y2 = a2
<=>
x2 y2
---- + ---- = 1
a2 b2
The points A(a,0) A'(-a,0) B(0,b) B'(0,-b) are called the vertices of the ellipse.
[AA'] is called the major axis and has 2a as length.
[BB'] is called the minor axis and has 2b as length.
On the previous figure we see that P'(a cos(t) , a sin(t))
By the definition of P we have P(a cos(t) , b sin(t))
From this it follows that
x = a cos(t)
y = b sin(t)
are parametric equations of the ellipse.
One can construct, for every t, a point on the ellipse.
Draw a line parallel to the y-axis through the intersection point of OP' and the circle with radius a.
Draw a line parallel to the x-axis through the intersection point of OP' and the circle with radius b.
The intersection point of these lines is a point of the ellipse.
We define c as c2 = a2 - b2 met c > 0 or c = 0
We define the points F(c,0) and F'(-c,0) as the foci of the ellipse.
Note that the distance from a focus to point B(0,b) is equal to a.
Say P(a cos(t) , b sin(t)) is a variable point of the ellipse.
|PF|2 = (a cos(t) - c)2 + b2 sin2(t)
= a2 cos2(t) - 2 a c cos(t) + c2 + (a2 - c2) sin2(t)
= a2 - 2 a c cos(t) + c2 cos2(t)
= (a - c cos(t))2
|PF'|2 = ... = (a + c cos(t))2
Since a > c :
|PF| + |PF'| = a - c cos(t) + a + c cos(t) = 2a = constant
Conversely, we show that point P is on the ellipse if |PF| + |PF'| = 2a.
Connect P with F and F'. Say P' in the intersection point of F'P with the ellipse. (see figure)
If P is not on the ellipse then
|PF| + |PF'| = 2a
and |P'F| + |P'F'| = 2a
We have |PF| + |PF'| -|P'F| - |P'F'| = 0
Then, we have |P'P| + |PF| = |P'F|
and this is impossible.
Conclusion:
If F and F' are the foci of een ellipse x2/a2 + y2/b2 = 1 , then we have
for each point P of the ellipse |PF| + |PF'| = 2a.
To obtain the slope of the tangent line we differentiate implicitly.
2x 2y y'
-- + ----- = 0
a2 b2
Solving for y', we obtain
b2 x
y'= - ----
a2 y
Say D(xo,yo) is a fixed point of the ellipse.
The slope of the tangent line in point D is
b2 xo
y'= - ------
a2 yo
The equation of the tangent line is
b2 xo
y - yo = - ----- (x - xo)
a2 yo
<=>
a2 yo y - a2 yo2 = b2 xo2 - b2 xo x
<=>
a2 yo y + b2 xo x = a2 yo2 + b2 xo2
<=>
since D(xo,yo) is on the ellipse
a2 yo y + b2 xo x = a2 b2
<=>
xo x yo y
---- + ---- = 1
a2 b2
The last equation is the tangent line in point D(xo,yo) of an ellipse.
Take the bisectors t and n of the lines DF and DF'.
Say F" is the reflection point of F in t.
Take any point T on t different from D.
Since |D,F| = |D,F"| , |F',F"| = 2a .
Now in the triangle F'TF" , we see that
|F',T| + |T,F"| > 2a
=> |T,F'| + |T,F| > 2a
And from the definition of ellipse, it follows that T is outside of the
ellipse. Hence all the points of t, different from D, are outside of the
ellipse and therefore the bisector t of the lines DF and DF' is a
tangent line of the ellipse.
The line n is a normal of the ellipse.
Since |F',F"| = 2a = constant, we see that the mirror image of F with
respect to a variable tangent line is on the circle with center F' and with
radius 2a.
Call P the projection of F on the tangent line.
Point O is the midpoint of the segment [F,F'] and point P is midpoint
of the segment [F,F"]. Hence |O,P| = a .
The orthogonal projection of F on a variable tangent line is
the circle with center O and radius a.
The given solution is not 'the' solution.
Most exercises can be solved in different ways.
It is strongly recommended that you at least try to solve the
problem before you read the solution.
Find the equation of the normal line in a point P(xo,yo) of the ellipse
b2 x2 + a2 y2 = a2 b2
We know that the equation of the tangent line in P(xo,yo) is
a2 yo y + b2 xo x = a2 b2
The slope is
b2 xo
- -----
a2 yo
The slope of the normal line is
a2 yo
-----
b2 xo
The normal line is
a2 yo
y - yo = ------- (x - xo)
b2 xo
Take on the ellipse E a variable point P and F is F(c,0).
Show that the following lines are concurrent.
The tangent line to the ellipse in P
The perpendicular to PF in F
The line x = a2/c (directrix associated to F)
P( a cos(t), b sin(t) )
The tangent line in P : x.cos(t)/a + y.sin(t)/b = 1 <=> xb cos(t) + ya sin(t) - ab = 0
The slope of PF is b sin(t) / ( a cos(t) -c )
The perpendicular to PF in F is
y.b sin(t) = (c - a cos(t))(x-c) <=> y b sin(t) + (a cos(t) -c) x + c (c - a cos(t)) = 0
The directrix associated to F : c x - a2 = 0
The concurrent condition is
| b cos(t) a sin(t) -ab |
| a cos(t) -c b sin(t) c(c-a cos(t)) | = 0
| c 0 -a2 |
We unfold the determinant following row 3 and taking into account that a2 = b2 + c2.
After simplification the cofactor of c is a2.sin(t)(a-c cos(t))
After simplification the cofactor of -a2 is c.sin(t).(a-c cos(t))
From this it follows that the determinant is 0.
P is a variable point on the ellipse E and F is F(c,0).
d is the line x = a2/c (directrix associated to F).
Show that |PF|/|P,d| is constant.
P( a cos(t), b sin(t) )
|P,d|2 = (a2/c - a cos(t))2 = (a2/c2).(a - c cos(t))2.
|PF|2 = (a cos(t) -c)2 + (a2-c2)sin2(t) = ... = (a - c cos(t))2.
From this it follows that |PF|/|P,d| is constant.
This constant is called the eccentricity of the ellipse.
Find the product of the distances from the foci of an ellipse to a fixed tangent line.
Show that this product is constant.
Take an ellipse E
b2 x2 + a2 y2 = a2 b2
and a point P on E
P(a cos(t) , b sin(t))
The tangent line is
a2 yo y + b2 xo x = a2 b2
<=>
a2 b sin(t) y + b2 a cos(t) x = a2 b2
<=>
a sin(t) y + b cos(t) x = a b
The normal equation of the tangent line is
a sin(t) y + b cos(t) x - a b
------------------------------ = 0
_________________________
| 2 2 2 2
\| a sin (t) + b cos (t)
The distance from F(c,0) to the tangent line is
b cos(t) c - a b
| ------------------------------|
_________________________
| 2 2 2 2
\| a sin (t) + b cos (t)
The distance from F'(-c,0) to the tangent line is
- b cos(t) c - a b
| ------------------------------|
_________________________
| 2 2 2 2
\| a sin (t) + b cos (t)
The product is
a2 b2 - b2 c2 cos2(t)
| ---------------------------|
a2 sin2(t) + b2 cos2(t)
b2(a2- c2 cos2(t))
= | ------------------------------ |
a2 (1 - cos2(t)) + b2 cos2(t)
b2(a2- c2 cos2(t))
= | --------------------|
a2 - c2 cos2(t)
= b2
In an ellipse E take all chords with slope m.
Show that the centers of these chords are on one line.
2 2
x y
-- + -- = 1
2 2
a b
<=>
b2 x2 + a2 y2 = a2 b2
A variable line with slope m is y = m x + t
The intersection points of E and the line are the solutions of
the system
/
| b2 x2 + a2 y2 = a2 b2
|
\ y = m x + t
Substitution gives
b2 x2 + a2 (m x + t)2 - a2 b2 = 0
<=>
(b2 + m2 a2) x2 + 2 a2 t m x + t2 a2 - b2 a2 = 0
Say x1 and x2 are the roots of this equation.
The abscissae of the center points are
x1 + x2 - a2 t m
------- = -----------
2 b2 + m2 a2
The centers of the chords are the intersection points of
- a2 t m
x = ------------ and y = m x + t
b2 + m2 a2
P(x,y) is the center of a chord
<=>
both equations have a common t-value
The condition arises if we eliminate t from these 2 equations.
We obtain the locus of the intersection points.
m y a2 + x b2= 0
<=>
b2 x
y = - ----
a2 m
P is a variable point of the ellipse b2 x2 + a2 y2 =a2 b2.
Let A = A(a,0) and A' = A'(-a,0).
M1 is on the x-axis such that APM1 is a right angled triangle in P.
M2 is on the x-axis such that A'PM2 is a right angled triangle in P.
Show that the distance |M1 M2| is constant.
P( a cos(t), b sin(t) )
b sin(t)
slope of PA = --------------
a(cos(t) -1)
a (1 - cos(t))
slope of PM1 = -----------------
b sin(t)
line PM1 is
a (1 - cos(t))
y - b sin(t) = ---------------- ( x - a cos(t))
b sin(t)
- b2 sin2(t)
M1 ( ------------------ + a cos(t) , 0)
a (1 - cos(t))
In the same way, we calculate M2
b2 sin2(t)
M2 ( ------------------ + a cos(t) , 0)
a (1 + cos(t))
|M1 M2| = ... = (2 b2)/a = constant
A variable tangent line to the standard ellipse forms with the x-axis and the y-axis
a triangle. Find the tangent lines such that the area of the triangle is minimum.
Key ideas:
Because of the symmetry it is sufficient to work in the first quadrant.
We work with the equation of the tangent line at a variable point.
The equation of the tangent line at a variable point (a cos(t) , b sin(t)) is
x cos(t)/a + y sin(t)/b = 1
Intersection with x-axis is ( a/cos(t) , 0 )
Intersection with y-axis is ( 0 , b/sin(t) )
Area of the triangle is :
a b ab
--------------- = ---------
2 sin(t) cos(t) sin(2t)
The area is minimum if and only if sin(2t) = 1.
So, for t = pi/4 the area is minimum.
Conclusion: the equation of the tangent line is x/a + y/b = sqrt(2)
and the reflections of that line relative to the x-axis and y-axis..
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