Take a line d and a point F not on d.
The locus of all points P such that |P,d| = |P,F| is a parabola.
To obtain an equation, we choose the x-axis and y-axis as in the
figure below.
We give F coordinates (p/2,0).
Then we have d with equation x = - p/2.
P(x,y) is on the parabola
<=>
|P,d| = |P,F|
<=>
|P,d|^{2} = |P,F|^{2}
<=>
p 2 p 2
(x + -) = (x - -) + y^{2}
2 2
<=>
...
<=>
y^{2} = 2p x
The point F is called the focus and the line d is the directrix.
The axis of symmetry of the parabola is called the axis of the parabola.
The axis of the parabola is the line through the focus and perpendicular to the directrix. The intersection point of that axis and the parabola is called the vertex of the parabola.
x^{2} = 2 p y
<=>
1
y = ----- x^{2}
2 p
Now the focus is (0, p/2) and the directrix is y = -p/2.
Let a = 1/ (2 p)
Now the equation is
y = a x^{2}
The focus is ( 0, 1/(4a) ) and the directrix is y = - 1/(4a)
Since the equation is y = ax^{2} , we can find the slope of the tangent line by means
of the derivative. In point P(x_{o}, a x_{o}^{2}) of the parabola the slope of the
tangent line is 2 a x_{o}.
A translation of the parabola y = a x^{2} gives a parabola with an equation of the form
y = ax^{2} + bx + c
The equation y^{2} = 2p x owes its simplicity to the special location of the directrix
and the focus relative to the axes.
However, the parabola was defined as a set of points which satisfy a geometric condition.
Now, we set up the equation of a parabola with focus F(a,b) and directrix ux+vy+w=0.
Point P(x,y) is on the parabola if and only if |P,d| = |P,F|.
Since we need the distance from the line ux+vy+w=0 , we first bring this equation in its normal form
l x + m y + n = 0 with l^{2} + m^{2} =1.
P(x,y) is on the parabola
<=>
|P,F| = |P,d|
<=>
|P,F|^{2} = |P,d|^{2}
<=>
(x - a)^{2} + (y - b)^{2} = (l x + m y + n)^{2} (1)
Example:
We calculate the equation of the parabola with focus F(1,2) and directrix d: 3x+4y-2=0.
The directrix has normal equation (3/5) x + (4/5)y - 2/5 = 0.
The equation of the parabola is
16 x^{2} - 24 xy + 9 y^{2} - 38 x - 84 y + 121 = 0 (3)
The annoying thing is that that form not shows where the focus is and where te directrix is.
If we want F and d, we have to calculate F en d from the last expression.
Therefore we bring (1) in another shape.
16 x^{2} - 24 xy + 9 y^{2} - 38 x - 84 y + 121 = 0
<=>
(4 x - 3 y)^{2} = 38 x + 84 y - 121
Since l^{2} + m^{2} must be equal to 1, we divide both sides by 25
<=>
( (4/5) x - (3/5) y)^{2} = 38/25 x + 84/25 y - 121/25
We compare this result with (4) and we see that m = (4/5) en l = (3/5)
a + l.n = 19/25 (5)
b + m.n = 42/25 (6)
n^{2} - a^{2} - b^{2} = - 121/25 (7)
Since l and m is known, we bring a and b from (5) and (6) to (7) and then
a = 1 ; b = 2 ; n = -2/5.
The equation (3) is a parabola with focus F(1,2) and directrix
(3/5)x + (4/5)y -2/5 = 0 <=> 3x+4y-2=0.
Focus in O(0,0)
x^{2} +y^{2} = (l x + m y + n)^{2}
Focus in O(0,0) and the directrix parallel to the y-axis.
x^{2} +y^{2} = (x-c)^{2}
Consider c as a parameter. Then we have a set of parabolas with a fixed focus and a fixed axis.
Such a set of parabolas is called confocal parabolas.
Focus in O(0,0) and the directrix parallel to the x-axis.
x^{2} +y^{2} = (y-c)^{2}
Again we have a set confocal parabolas. The parameter is c.
We start with a fixed coordinate system, an arbitrary parabola P and a fixed
parabola P" with equation y = x^{2}.
There is always a suitable rotation and translation such that P is transformed in
a parabola P' with equation y = ax^{2} with a > 0.
So, P is similar to P'.
We also know that a homothetic transformation transforms a figure in a similar figure.
We choose O(0,0) as center of a homothety h and we choose 'a' as factor.
The transformation formulas are
h
(x,y) ----> (ax, ay)
Point D' is on parabola P'
<=> D'(x , a x^{2})
h transforms this point D' in the point D"
D"( a (x) , a (a x^{2})) = D"( (a x) , (a x)^{2} )
<=> Point D" is on the parabola P" with equation y = x^{2}
So, the arbitrary parabola P is similar to the fixed parabola P".
Thus all parabolas are similar to the same fixed parabola.
This means that all parabolas are similar.
The shape of the parabola is determined by a geometric property.
It does not depends on the chosen coordinate system.
A parabola is completely determined by a given focus and directrix.
Consider two parabolas :
P_{1} with focus F_{1} and directrix d_{1}
and
P_{2} with focus F_{2} and directrix d_{2}
The parabolas P_{1} and P_{2} are congruent
<=>
There is a displacement v such that v(F_{1}) = F_{2} and v(d_{1}) = d_{2}
<=>
The relative position of F_{1} against d_{1}
is the same as
The relative position of F_{2} against d_{2}
<=>
The distance from F_{1} to d_{1} = the distance from F_{2} to d_{2}
<=>
The distance from F_{1} to the tangent at the vertex of P_{1}
is equal to
The distance from F_{2} to the tangent at the vertex of P_{2}
Example:
The parabola P_{1} with equation y = 0.25 x^{2} is congruent with the
parabola P_{2} with directrix d_{2} : x - y = 0.
The focus of P_{2} is on the line b with equation x + y - 1 = 0.
Find all possible equations of P_{2}.
The focus of P_{1} is (0,1) and the directrix is y+1=0.
This focus lies at a distance 2 from the directrix.
The parabola P_{2} is congruent with P_{1} if and only if
the focus F_{2} lies at a distance 2 from the directrix x - y = 0.
Let F_{2} be a variable point of the line b. We use parameter t.
F_{2} = F_{2}(t , 1-t).
Take in a plane two lines a and b with resp. equations
x = 2 p t^{2} (1)
y = 2 p t (2)
The real number t is the parameter.
We know, from the theory of 'Elimination of parameters', that the
intersection points of the two associated lines constitute a curve.
To obtain the equation of that curve, we eliminate the parameter t
from the two equations. This means that we search for the condition
such that (1) and (2) has a solution for t.
From (2) we have t = y / (2p).
So, this t-value is a solution of (1) if and only if
y 2
x = 2 p (---) <=> y^{2} = 2 p x
2 p
Hence, the two associated lines constitute a curve and that curve
is the parabola.
We say that (1) and (2) are parametric equations of the parabola.
The point
D( 2 p t^{2} , 2 p t)
is on the parabola for each t-value and with each point of the parabola
corresponds a t-value.
To obtain the slope of the tangent line we differentiate implicitly.
2 y y' = 2 p
<=>
y' = p/y
Say D(x_{o},y_{o}) is a fixed point of the parabola.
The slope of the tangent line in point D is
p
---
y_{0}
The equation of the tangent line is
p
y - y_{0} = -- (x - x_{0})
y_{0}
<=>
y_{0} y - y_{0}^{2} = p x - p x_{0}
Since y_{0}^{2} = 2p x_{0}
<=>
y_{0} y - 2 p x_{0} = p x - p x_{0}
<=>
y y_{0} = p (x + x_{0})
The last equation is the tangent line in point D(x_{0},y_{0}) of a parabola.
It is easy to show that this line meets the x-axis at the point s(-x_{0},0).
From this it is easy to construct the tangent line in a
given point D. (see figure)
|C,D| = |D,F| = |E,F| = x_{0} + p/2 and so CDEF is a rhomb.
Hence the tangent line bisects the angle CDF.
Point C is the mirror image of F relative to the tangent line.
So, the mirror image of F relative to a variable tangent line is
the directrix.
Additionally, the orthogonal projection of F on a variable tangent line is
the tangent line through the vertex of the parabola.
The line through point D and orthogonal with the tangent line is called
the normal at point D.
The normal through D is also a bisecting line of CD and DF. D_{1} = D_{2}.
From this property it follows that parabolas have a certain reflection property.
All light rays entering parallel to the axis of a parabola are reflected to the focus.
This works in the other direction too. A point source placed at the focus will send
light out in a collimated beam parallel to the axis.
Take a line t with a given slope m. The equation is y = m x + q.
The intersection points with the parabola are the solutions
of the system
y^{2} = 2 p x
y = m x + q
Substitution gives
(m x + q)^{2} = 2 p x
<=>
m^{2} x^{2} + 2 (m q - p) x + q^{2} = 0
The line t is a tangent line if and only if the roots of the last
equation are equal. Therefore the discriminant has to be zero.
4 (m q - p)^{2} - 4 m^{2} q = 0
<=>
4 p (p - 2 m q) = 0
<=>
p
q = ---
2 m
The tangent line with a given slope m is
p
y = m x + ---
2 m
Take a fixed point P(x_{0},y_{0}) .
We'll calculate the tangent lines from P to the parabola .
A line t with variable slope through P is
y - y_{0} = m(x - x_{0})
The intersection points with the parabola are the solutions
of the system
y^{2} = 2 p x
y - y_{0} = m(x - x_{0})
We substitute x from the first equation into the second one.
y^{2}
y - y_{0} = m(--- - x_{0})
2 p
<=>
- m y^{2} + 2 p y - 2 p y_{0} + 2 p m x_{0} = 0
The line t is a tangent line if and only if the roots of the last
equation (in y) are equal. Therefore the discriminant has to be zero.
4 p^{2} + 4 m (2 p m x_{0} - 2 p y_{0}) = 0
<=>
2 x_{0} m^{2} - 2 y_{0} m + p = 0
The roots of this equation are the slopes of the two tangent lines.
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 points of the parabola y^{2} = 4x such that the distance from these points to the focus is four.
First method
The focus is F(1,0).
The required points are the intersections of the parabola with a circle (x-1)^{2} + y^{2} = 16.
We solve the system formed by the equation of the parabola and the equation of the circle.
We find two points : (3, ± sqrt(12)).
Second method
The points at a distance 4 from the focus, are also at a distance 4 of the directive.
The directive has an equation x = -1. The required points are on the line x = 3.
So they have an abscissa 3. We find (3, ± sqrt(12)).
The parabola P has equation y^{2} = 2 p x.
A variable point P has coordinates (p/2,t). The parameter is
the real number t greater than p .
Calculate the the tangent of the sharp angle between
the tangent lines through P.
From the theory about the tangent lines through a given point P(x_{o},y_{o}),
we know that the slopes are given by
______________
| 2
yo + \| yo - 2 p xo
m_{1} = ---------------------
2 xo
______________
| 2
yo - \| yo - 2 p xo
m_{2} = ---------------------
2 xo
In our case here, xo = p/2 and yo = t. So, the slopes are
______________
| 2 2
t + \| t - p
m_{1} = ---------------------
p
______________
| 2 2
t - \| t - p
m_{2} = ---------------------
p
The tangent of the angle between the lines is given by
m_{1} - m_{2}
-----------
1 + m_{1} m_{2}
Now,
__________
| 2 2
2 \| t - p
m_{1} - m_{2} = -------------- and m_{1}.m_{2} = 1
p
Then the tangent of the angle between the lines is
__________
| 2 2
\| t - p
-------------
p
The tangent lines t and t' in points P(x_{1},y_{1}) and in P'(x_{2},y_{2}) of a parabola
are orthogonal.
Prove that y_{1}.y_{2} = - p.p
From the theory about the tangent line in a given point P(x_{o},y_{o})
of a parabola, we know that the slope is p/y_{o} .
Thus, in point P is the slope p/y_{1} and in point P' is the slope p/y_{2}.
The tangent lines are orthogonal if and only if
p p
-- . -- = -1
y_{1} y_{2}
<=> y_{1}.y_{2} = - p.p
Point P is on a parabola. The projection of P on the axis of the parabola is Q. The normal line in P
intersects the axis in point N. Show that |QN| is constant.
P( 2 p t^{2} , 2 p t) is a variable point on the parabola. Then Q is Q( 2 p t^{2} , 0).
The tangent line in P has slope p/(2pt) = 1/(2t) .
The slope of the normal line is -2t. The normal line in P is y - 2pt = -2t(x - 2pt^{2}).
The normal line in P intersects the x-axis in point N(p+2pt^{2},0). |QN| = p = constant.
The line y = x+3 is a tangent to the parabola y^{2} = 2px. Calculate p.
The tangent line in a variable point P( 2 p t^{2} , 2 p t) of the parabola is 2pt y = x + 2pt^{2}.
That line should coincide with the given line. It follows that p = 1/6.
The line y = 0.5 x - 4 is a normal line to the parabola y^{2} = 2px.
Find the tangent line to the parabola corresponding with that normal line.
The tangent line is orthogonal to the normal line and so the slope is -2.
The tangent line with that direction is y = -2 x - p/4
We consider two points P and P' on a parabola. The tangent lines in these points intersect on the directrix.
Find the product of the ordinates of P and P'.
Take P( 2 p t^{2} , 2 p t) and P'( 2 p t'^{2} , 2 p t') on the parabola.
We calculate the tangent lines in P and P'. We find x - 2ty + 2pt^{2} = 0 and x - 2t'y + 2pt'^{2} = 0.
We require that the two tangent lines and the directrix are concurrent lines. The condition is:
Consider the tangent lines in 4 points of a parabola. They intersect and form a quadrilateral.
Show that the line through the midpoints of the diagonals of the quadrilateral is parallel to the axis of the parabola.
For i = 1,2,3,4 is:
We take 4 points P_{i}(2pt_{i}^{2}, 2pt_{i}) on the parabola.
The four tangent lines l_{i} are x - 2 t_{i} y + 2 p t_{i}^{2} = 0.
The ordinate of the intersection point of l_{1} and l_{2} is p.(t_{1} + t_{2})
So, the ordinate of the intersection point of l_{i} and l_{j} is p.(t_{i} + t_{j}) .
We calculate the y-value of the center of each diagonal. We find (1/2).p.(t_{1}+t_{2}+t_{3}+t_{4}).
Since the two centers have the same ordinate, the line through these centers is parallel
to the axis of the parabola.
Take four points on a parabola. Find the condition in order that the four point are concyclic.
For i = 1,2,3,4 is:
We take 4 points P_{i}(2pt_{i}^{2}, 2pt_{i}) on the parabola.
From the theory of the circle we know that the 4 points are concyclic if and only if
We know that x^{2} + y^{2} = (x - c)^{2} is the equation of a set confocal parabolas with focus F(0,0).
Q(r,s) is a point different from F. Show that there are exactly two parabolas of the set through Q.
Show that there is an orthogonal intersection of these two parabolas in point Q.
Part 1:
For each c-value there is a parabola in the set. We'll find the two c-values such that Q(r,s) is
on a parabola.
Q is on a parabola of the set
<=>
r^{2} + s^{2} = (r - c)^{2}
<=>
...
<=>
c^{2} - 2 r c - s^{2} = 0 (*)
The discriminant of the quadratic equation is 4(r^{2}+ s^{2}). Since Q is different from F, D is > 0.
So, there are two values for c such that Q is on the parabola. There are 2 parabolas through Q.
Part 2:
Say c_{1} and c_{2} are the two roots of (*). First, we take the parabola corresponding with c = c_{1}.
x^{2} + y^{2} = (x - c_{1})^{2}
<=>
y^{2} = - 2 c_{1} x + c_{1}^{2}
We calculate the derivative by Implicit differentiation
2 y y' = - 2 c_{1}
<=>
y' = - c_{1}/y
In Q(r,s) the slope of the tangent line is - c_{1}/s
In the same way we find for c = c_{2}.
In Q(r,s) the slope of the tangent line is - c_{2}/s
The product of these slopes is c_{1}.c_{2}/s^{2}
But from (*) we see that the product of the roots is c_{1}c_{2} = -s^{2}.
From this, the product of the slopes is c_{1}.c_{2}/s^{2} = -1.
So, the tangent lines are orthogonal in Q.
Consider a variable tangent line to the parabola y^{2} = 2 p x.
This line cuts the x-axis at point Q and the y-axis at point R.
Find the locus of the center M of [QR].
A point P(x_{o}, y_{o}) is on a parabola.
x_{o} and y_{o} are parameters connected by the binding equation y_{o}^{2} = 2 p x_{o}.
The tangent line in a random point P(x_{o},y_{o}) is y y_{o} = p (x + x_{o}).
The intersection points with the axes are Q(-x_{o} , 0) and R(0, px_{o}/y_{o} ).
The center M is M(-x_{o}/2 , px_{o}/ (2y_{o}) ).
M is the intersection point of the associated lines x = -x_{o}/2 and y = px_{o}/ (2y_{o}).
To find the locus, we eliminate x_{o} and y_{o} between the associated lines and the
binding equation.
/ x = -x_{o}/2
| y = px_{o}/ (2y_{o})
\ y_{o}^{2} = 2 p x_{o}
We calculate x_{o} and y_{o} from the first two equations and we put the results in the last one. The equation of the locus is y^{2} = - p x/4.
A variable line l has a fixed slope m. The line l intersects the parabola y^{2} = 2px
in the points Q and R.
Show that the center M of [QR] lies on a fixed line parallel to the axis of the parabola.
De line l has an equation y = mx + q. q is a parameter.
The points Q and R are the solutions of the system
/ y = m x + q
\ y^{2} = 2 p x
<=>
/ x = (y-q)/m
\ y^{2} = 2p (y-q)/m
<=>
/ x = (y-q)/m
\ y^{2} - 2py/m + 2pq/m = 0 (*)
We call y_{1} and y_{2} the solutions of the quadratic equation (*).
y_{1} and y_{2} are the ordinates of Q and R.
The ordinate of the center M is (y_{1} + y_{2})/2 = half the sum of the roots of (*).
So, the ordinate of the center M is p/m. This is a fixed value as it is independent of q.
M is on a fixed line parallel to the axis of the parabola.
P is the parabola y^{2} = 2p x. We connect a variable point P of the parabola
with the vertex of the parabola.
Show that the locus of the center M of this variable chord PO is a parabola.
The transformation which transforms P into M is actually a homothetic transformation h
with center O and factor 1/2. The image of P through h is a figure that is
similar to the parabola P.
Therefore, this figure is a parabola.
A parabola has an equation x^{2} + y^{2} = (x+1)^{2}. Find the focus F and the directrix.
A variable line through F intersects the parabola in the points A and B.
Find the locus of the center M of [AB].
The focus is F(0,0) and the directrix is x = -1.
The equation of the parabola can be simplified to y^{2} = 2x + 1.
The variable line through F is y = m x. m is parameter.
The intersection points A and B are the solutions of the system
y^{2} = 2x + 1 (1)
y = m x (2)
If we replace y in (1) by m x, then we get m^{2} x^{2} - 2 x -1 = 0
The solutions of this equation are the x-values x_{1} and x_{2} of A and B.
The center of [AB] has coordinates
x = (x_{1} + x_{2})/2 = (2/m^{2}) / 2 = 1/m^{2}
y = m x = 1/m
So, M( 1/m^{2}, 1/m )
The coordinates of M tell us, without calculation, that M moves along the curve
x=y^{2}.
The locus of M is the parabola y^{2}= x.
Given
Parabola with equation y = 0.25 x^{2}
Points A(7,0) and B(3/2,-1)
line r with equation y = 2 x - 4
Find:
Focus and directrix ; make a figure
The two tangent lines through point A
The tangent line perpendicular to the line r
The directrix is y = -1 and F(0,1)
It is clear that the first tangent line through A is y = 0.
Now, we retake the general figure with many properties of the parabola.
We see that: if point S is on the tangent line through the vertex then
the second tangent line through S is perpendicular to SF.
In our exercise, the second tangent line through A is perpendicular to AF.
The slope of AF is -1/7. So, the required tangent line has a slope = 7.
The second tangent line is y = 7(x - 7).
Since r and r' are two perpendicular tangent lines, they intersect on the directrix.
The intersection point of r with the directrix is R(1.5 , -1).
The tangent line through R and perpendicular to r is
y + 1 = - 0.5 (x - 1.5) <=> y = -0.5 x - 0.25
A is a variable point of the parabola y^{2}=6x and B(2,8) is a fixed point.
Find A such that |AB| is minimum.
We use the parametric equations of the parabola. Take A(6t^{2},6t) with t as parameter.
|AB| is minimum if and only if |AB|^{2} is minimum
|AB|^{2} = (6t^{2} - 2)^{2} + (6t - 8)^{2}
= 36 t^{4} + 12 t^{2} - 96 t + 68
|AB| is minimum if and only if the derivative is zero.
<=> 12 t^{3} + 2 t - 8 = 0
The points A and B are on the parabola y^{2} = 2p x. M is the center of the segment [A,B].
The lines r an r' are the tangent lines in A and B to the parabola. The line s is parallel to
the axis of the parabola and contains M. Show that the lines r, r' and s are concurrent lines.
Let A(2pt^{2} , 2pt) , B(2pt'^{2} , 2pt')
Then M(... , p(t+t'))
line r : y 2pt = p ( x + 2pt^{2}) <=> x - 2 t y + 2 p t^{2} = 0
line r': y 2pt' = p ( x + 2pt'^{2}) <=> x - 2 t' y + 2 p t'^{2} = 0
line s : y = p(t+t') <=> 0x + y - p (t+t') = 0
The lines are concurrent if and only if
| 1 -2t 2pt^{2} |
| 1 -2t' 2pt'^{2} | = 0
| 0 1 -p(t+t')|
Row 2 - Row 1
<=>
| 1 -2t 2pt^{2} |
| 0 2(t-t') 2p(t'^{2}-t^{2}) | = 0
| 0 1 -p(t+t') |
<=>
-2p(t^{2}-t'^{2}) - 2p(t'^{2}-t^{2}) = 0
Find t such that the following parabolas are congruent. P_{1} : y^{2} = t x with t > 0 P_{2} : (x - 1)^{2} + (y - 2)^{2} = (0.6 x + 0.8 y)^{2}
The parabola P_{1} has focus F_{1}(t/4, 0) and directrix d_{1}: x = - t/4.
The distance from F_{1} to d_{1} is t/2.
The parabola P_{2} has focus F_{2}(1,2) and directrix d_{2}: 0.6 x + 0.8 y = 0.
The distance from F_{2} to d_{2} is 2.2.
So, t = 4.4
The parabola y^{2} = 4.4 x^{2} is congruent with (x - 1)^{2} + (y - 2)^{2} = (0.6 x + 0.8 y)^{2}.
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