The Parabola

Definition and equation

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 - -)  + y2
2          2
<=>

...

<=>

y2  = 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.

Other forms of the equation of a parabola

1. We swap the x-axis and y-axis

Then we have a parabola with an equation

```
x2 = 2 p y
<=>
1
y = ----- x2
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 x2

```
The focus is ( 0, 1/(4a) ) and the directrix is y = - 1/(4a)

Since the equation is y = ax2 , we can find the slope of the tangent line by means of the derivative. In point P(xo, a xo2) of the parabola the slope of the tangent line is 2 a xo.

2. A translation of the parabola y = a x2 gives a parabola with an equation of the form
y = ax2 + bx + c

3. The equation y2 = 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 l2 + m2 =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

```
(x-1)2 + (y-2)2 = ((3/5) x + (4/5)y - 2/5)2             (2)
```
It is obvious that this form can be simplified to
```
16 x2 - 24 xy + 9 y2 - 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.
```
(x - a)2 + (y - b)2 = (l x + m y + n)2
<=>
x2 - 2ax + a2 + y2 -2by + b2 = l2 x2 + m2 y2 + n2  + 2 l.m xy + 2 l.n x + 2m.n y
<=>
(1-l2)x2 -  2 lm xy + (1-m2) y2 = 2x(a + l.n) + 2y(b + m.n) -a2 - b2 + n2
<=>
(m x - l y)2 = 2x(a + l.n) + 2y(b + m.n) -a2 - b2 + n2    (4)
```
We transform (3) to the shape (2).
```
16 x2 - 24 xy + 9 y2 - 38 x - 84 y + 121 = 0
<=>
(4 x - 3 y)2 =  38 x + 84 y - 121

Since l2 + m2 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)
n2 - a2 - b2 = - 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.
```
4. Focus in O(0,0)
x2 +y2 = (l x + m y + n)2

5. Focus in O(0,0) and the directrix parallel to the y-axis.
x2 +y2 = (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.

6. Focus in O(0,0) and the directrix parallel to the x-axis.
x2 +y2 = (y-c)2
Again we have a set confocal parabolas. The parameter is c.
```

```

Similar parabolas

We"ll show that all parabolas are similar.

We start with a fixed coordinate system, an arbitrary parabola P and a fixed parabola P" with equation y = x2.

There is always a suitable rotation and translation such that P is transformed in a parabola P' with equation y = ax2 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 x2)

h transforms this point D' in the point D"

D"( a (x) , a (a x2)) = D"( (a x) , (a x)2 )

<=> Point D" is on the parabola P" with equation  y = x2

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.

Congruent parabolas

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 :

```
P1 with focus F1 and directrix d1
and
P2 with focus F2 and directrix d2

The parabolas P1 and P2 are congruent

<=>

There is a displacement v such that   v(F1) = F2 and v(d1) = d2

<=>

The relative position of F1 against d1
is the same as
The relative position of F2 against d2

<=>

The distance from  F1 to d1 = the distance from F2 to d2

<=>

The distance from  F1 to the tangent at the vertex of P1
is equal to
The distance from  F2 to the tangent at the vertex of P2
```
Example:
 The parabola P1 with equation y = 0.25 x2 is congruent with the parabola P2 with directrix d2 : x - y = 0. The focus of P2 is on the line b with equation x + y - 1 = 0. Find all possible equations of P2.

The focus of P1 is (0,1) and the directrix is y+1=0. This focus lies at a distance 2 from the directrix.

The parabola P2 is congruent with P1 if and only if the focus F2 lies at a distance 2 from the directrix x - y = 0.

Let F2 be a variable point of the line b. We use parameter t.
F2 = F2(t , 1-t).

```
| F2, d2| = 2
<=>
| 2 t -1|
------------- = 2
sqrt(2)
<=>
(2t - 1)2 = 8
<=>
t = 0.5 ± sqrt(2)
```
So there are two possible locations for the F2.
1. t = 0.5 + sqrt(2)
```
F2 = F2(0.5 + sqrt(2) , 0.5 -sqrt(2) )

The equation of the parabola P2 is

(x - 0.5 -sqrt(2))2  + (y - 0.5 + sqrt(2))2 = (x-y)2/2

<=>

x2 + 2 x y + y2 - 7.66 x + 3.66 y + 9 = 0

```
2. t = 0.5 - sqrt(2)

Solve this case as exercise

Parametric equations of the parabola

Take in a plane two lines a and b with resp. equations
```
x = 2 p t2             (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 (---)    <=>  y2  = 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 t2  , 2 p t)
```
is on the parabola for each t-value and with each point of the parabola corresponds a t-value.

Tangent line in a point D of a parabola

Take the parabola
```
y2  = 2p x
```
To obtain the slope of the tangent line we differentiate implicitly.
```
2 y y' = 2 p
<=>
y' = p/y
```
Say D(xo,yo) is a fixed point of the parabola.
The slope of the tangent line in point D is
```
p
---
y0
```
The equation of the tangent line is
```
p
y - y0 = --  (x - x0)
y0
<=>
y0 y - y02  = p x - p x0

Since   y02  = 2p x0

<=>
y0 y - 2 p x0 = p x - p x0
<=>
y y0 = p (x + x0)
```
The last equation is the tangent line in point D(x0,y0) of a parabola.
It is easy to show that this line meets the x-axis at the point s(-x0,0).
From this it is easy to construct the tangent line in a given point D. (see figure)
```

```

Powerful properties

From this we deduce many properties.
• |C,D| = |D,F| = |E,F| = x0 + 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. D1 = D2.
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.

Tangent line with a given slope

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
```
y2  = 2 p x

y = m x + q

Substitution gives
(m x + q)2  = 2 p x
<=>
m2  x2  + 2 (m q - p) x + q2  = 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 m2  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
```

Tangent lines from a given point

Take a fixed point P(x0,y0) .
We'll calculate the tangent lines from P to the parabola .
A line t with variable slope through P is
```
y - y0 = m(x - x0)
```
The intersection points with the parabola are the solutions of the system
```
y2  = 2 p x

y - y0 = m(x - x0)

We substitute x from the first equation into the second one.
y2
y - y0 = m(--- - x0)
2 p
<=>
- m y2  + 2 p y - 2 p y0 + 2 p m x0 = 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 p2  + 4 m (2 p m x0 - 2 p y0) = 0
<=>
2 x0 m2  - 2 y0 m + p = 0
```
The roots of this equation are the slopes of the two tangent lines.
```

y0 + sqrt(y02 - 2 p x0)
m1 = ------------------------------
2 x0

y0 - sqrt(y02 - 2 p x0)
m2 =  ----------------------------
2 x0
```
The equations of the tangent lines are
```
y2
y - y0= m1(----- - x0)       ;
p

y2
y - y0= m2(----- - x0)
p

```
The two lines are orthogonal if and only if m1 . m2 = -1
```
p
<=>     ---- = -1
2 x0

-p
<=>     x0 = ----
2
```
From this we see that if point P is on the directrix, the tangent lines are orthogonal.

 The directrix is the locus of the points P such that the tangent lines through P to the parabola are perpendicular.

Evolute of a parabola

See Evolute of a parabola

Exercises:

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 y2 = 4x such that the distance from these points to the focus is four.

•  The parabola P has equation y2 = 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.

•  The tangent lines t and t' in points P(x1,y1) and in P'(x2,y2) of a parabola are orthogonal. Prove that y1.y2 = - 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.

•  The line y = x+3 is a tangent to the parabola y2 = 2px. Calculate p.

•  The line y = 0.5 x - 4 is a normal line to the parabola y2 = 2px. Find the tangent line to the parabola corresponding with that normal line.

•  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'.

•  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.

•  Take four points on a parabola. Find the condition in order that the four point are concyclic.

•  We know that x2 + y2 = (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.

•  Consider a variable tangent line to the parabola y2 = 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].

You can find here theory and examples about loci.

•  A variable line l has a fixed slope m. The line l intersects the parabola y2 = 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.

•  P is the parabola y2 = 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.

•  A parabola has an equation x2 + y2 = (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].

•  Given Parabola with equation y = 0.25 x2 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

•  A is a variable point of the parabola y2=6x and B(2,8) is a fixed point. Find A such that |AB| is minimum.

•  The points A and B are on the parabola y2 = 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.

•  Find t such that the following parabolas are congruent. P1 : y2 = t x with t > 0 P2 : (x - 1)2 + (y - 2)2 = (0.6 x + 0.8 y)2

Topics and Problems