# Parabola segment

1. Jan 28, 2008

### jrm2002

I have a parabola.

Now, i need to get the length of aprt of a parabolic segment--that is a quarter of a parabola or 1/10 of a parabola, 3/4 of a parabola or anything

How is this possible?

2. Jan 28, 2008

### HallsofIvy

Staff Emeritus
?? What do you mean by "a quarter of a parabola or 1/10 of a parabola, 3/4 of a parabola"? A parabola has infinite length. If you have a parabola opening downward and are looking at the part above y= 0, please say so.

Last edited: Feb 2, 2008
3. Feb 2, 2008

### jrm2002

Sorry---I will put it more clearly,

I have a parabola whose directrix is not parallel to x axis.

The equation of this parabola is:

x^2 + 2xy + y^2 - 16y + 32 = 0

I know three points which lie on the parabola [start point, end point, vertex].

Now, I want the length of the parabola

4. Feb 2, 2008

### Gib Z

What exactly makes you think that equation yields a parabola?

5. Feb 2, 2008

### HallsofIvy

Staff Emeritus
Whatever makes him thinks so, he is right. The quadratic part is $x^2+ 2xy+ y^2= (x+ y)^2$ Let u= x+ y. Then y= u- x and the equation becomes, in terms of u, $u^2- 16(u- x)- 32= 0$ or $u^2- 16u- 16x- 32= 0$ and, finally, $x= (1/16)u^2- u- 2$. That is a parabola at 45 degrees to the coordinate axes. There is no simple formula for arclength of a parabola. You would have to integrate [tex]\sqrt{1+ ((1/8)u- 1)^2}du[/itex] between the u values of the endpoints.

6. Feb 2, 2008

### jrm2002

To explain in detail what I want:

Now,

I have a parabola whose 3 points (x , y) are defined as:

Point 1 : (-9, -1)

Point 2 : (-6,6) [VERTEX POINT]

Point 3: (4,-9)

I need to get the length of the parabola.

Now, the directrix of this parabola is not parallel to x axis.OK?

So, I introduce a transformed coordinate system in which the directrix
is parallel to an axis:

Now, always if I draw a tangent to the parabola at the vertex, it will be parallel to the directrix, right?

So, in the transformed coordinate system , the inclination of directrix would be:

tan(inverse) of 6/-6 [SEE THE VERTEX POINT].Right?

that is : -45 degrees

Now,

A transformed coordinate axis which
is rotated about the origin through an angle T is given by:

[ x' ] [ Cos T Sin T ][ x ] [ x Cos T + y Sin T ]
| | = | || | = | |
[ y' ] [ -Sin T Cos T ][ y ] [ -x Sin T + y Cos T ]

So, I get,

x ' = (x+y) (-cos 45)

y' = (y - x) (-cos45)

Right?

So, the three points with respect to the rotated coordinate axis would be:

Point 1 = ( -7.071 , -5.6568)

Point 2 = (0, 8.4852)

Point 3 = (3.5355 , 9.1923)

Solving the equation,

y = Ax^2 + Bx +C

I get,

A = 0.5091
B = 3.1999
C = -8.4852

y = 0.5091x^2 + 3.1999x +(-8.4852)

Right?

If i want to find the length of the parabolic arc I use:

1+ dy/dx^2)^1/2

and integrate between -7.071 and 3.5355.

But correct answer is : 26.603

Where am I going wrong?

7. Feb 2, 2008

### jrm2002

Now,

I have a parabola whose 3 points (x , y) are defined as:

Point 1 : (-9, -1)

Point 2 : (-6,6) [VERTEX POINT]

Point 3: (4,-9)

I need to get the length of the parabola.

Now, the directrix of this parabola is not parallel to x axis.OK?

So, I introduce a transformed coordinate system in which the directrix
is parallel to an axis:

Now, always if I draw a tangent to the parabola at the vertex, it will be parallel to the directrix, right?

So, in the transformed coordinate system , the inclination of directrix would be:

tan(inverse) of 6/-6 [SEE THE VERTEX POINT].Right?

that is : -45 degrees

Now,

A transformed coordinate axis which
is rotated about the origin through an angle T is given by:

[ x' ] [ Cos T Sin T ][ x ] [ x Cos T + y Sin T ]
| | = | || | = | |
[ y' ] [ -Sin T Cos T ][ y ] [ -x Sin T + y Cos T ]

So, I get,

x ' = (x+y) (-cos 45)

y' = (y - x) (-cos45)

Right?

So, the three points with respect to the rotated coordinate axis would be:

Point 1 = ( -7.071 , -5.6568)

Point 2 = (0, 8.4852)

Point 3 = (3.5355 , 9.1923)

Solving the equation,

y = Ax^2 + Bx +C

I get,

A = 0.5091
B = 3.1999
C = -8.4852

y = 0.5091x^2 + 3.1999x +(-8.4852)

Right?

If i want to find the length of the parabolic arc I use:

1+ dy/dx^2)^1/2

and integrate between -7.071 and 3.5355.

But correct answer is : 26.603

Where am I going wrong?

8. Feb 2, 2008

### jrm2002

To explain in detail what i want:

Now,

I have a parabola whose 3 points (x , y) are defined as:

Point 1 : (-9, -1)

Point 2 : (-6,6) [VERTEX POINT]

Point 3: (4,-9)

I need to get the length of the parabola.

Now, the directrix of this parabola is not parallel to x axis.OK?

So, I introduce a transformed coordinate system in which the directrix
is parallel to an axis:

Now, always if I draw a tangent to the parabola at the vertex, it will be parallel to the directrix, right?

So, in the transformed coordinate system , the inclination of directrix would be:

tan(inverse) of 6/-6 [SEE THE VERTEX POINT].Right?

that is : -45 degrees

Now,

A transformed coordinate axis which
is rotated about the origin through an angle T is given by:

[ x' ] [ Cos T Sin T ][ x ] [ x Cos T + y Sin T ]
| | = | || | = | |
[ y' ] [ -Sin T Cos T ][ y ] [ -x Sin T + y Cos T ]

So, I get,

x ' = (x+y) (-cos 45)

y' = (y - x) (-cos45)

Right?

So, the three points with respect to the rotated coordinate axis would be:

Point 1 = ( -7.071 , -5.6568)

Point 2 = (0, 8.4852)

Point 3 = (3.5355 , 9.1923)

Solving the equation,

y = Ax^2 + Bx +C

I get,

A = 0.5091
B = 3.1999
C = -8.4852

y = 0.5091x^2 + 3.1999x +(-8.4852)

Right?

If i want to find the length of the parabolic arc I use:

1+ dy/dx^2)^1/2

and integrate between -7.071 and 3.5355.