8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 1/14

Construct a Parabola

This is an applet to construct a parabola from its definition. If needed,Free graph paper is available.

Definition: A parabola is the set of all points M in a plane such that thedistance from M to a fixed point F, the focus, is equal to the distancefrom M to a fixed line called the directrix.

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 2/14

The axis of the parabola is the line through F and perpendicular to the directrix.Point V on the axis and halfway between the focus F and the directrix is called thevertex. It can be shown that the equation of a parabola with a horizontal axis andopening towards increasing x values is given by:

y

2

= 4ax

where a is a constant. For a deeper understanding of the equation of a parabola Gohere.Follow the steps in the tutorial below to construct a parabola using the abovedefinition. The equation is used to verify the construction of the parabola. Examplesof applications of the parabolic shape as Parabolic Reflectors and Antennas areincluded.

Your browser is completely ignoring the <APPLET> tag!

TUTORIAL 1 - click on the button above "click here to start" to start the applet andMAXIMIZE the window obtained. The slider in the top left panel can be used tochange the value of a, do not use it now, a=1.

2 - Before you start the construction, note the following.

The directrix: vertical line at x = -1.The vertex: point V at (0,0).The focus: point F at (1,0).A marker: point M at (1,2), this point can be moved around freely.Point D on the directrix and has the same y-coordinate as point M.The distances d(F,M) and d(D,M) are both equal to 2. (their values and thecoordinates of points M and D are displayed on the top left of the main panel).

3 - Construction:a - Start by pressing the button "Plot Points", this will plot point M (in blue) as anelement of the set of points whose distances from the directrix and the focus areequal.b - Drag point M horizontally to a new position. Note the distance d(D,M).c - Now drag point M vertically untill the distances d(D,M) and d(F,M) are equal or close in value.d - Now press the button "Plot Points" to plot this point.

4 - Drag point M to a new position and repeat step 3 to plot another point.

5 - Repeat step 4 as many times as you can to plot points whose distances from thevertex and the focus are equal.

6 - You may also want to plot points in quadrant IV.

7 - Once you have enough points plotted, press the button "Plot/Delete Parabola" toplot the whole parabola in order to verify that all points whose distances from thevertex and the focus are equal, can be described by one single equation givenabove with a = 1. You may also want to drag point M along the parabola and see

that the distance from M to F is equal (or very close) to the distance from M to thedirectrix.

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 3/14

Geometry of the Parabola (2D)

 by Henri Picciotto

Parabolas are a central topic in high school algebra classes, but, perhaps because of the

rigid separation between algebra and geometry classes in the US secondary curriculum,

we do not usually treat them as geometric objects. While most teachers are aware of some

of the parabola's geometric properties, few of us are familiar with the proofs of those properties.

On this page, I present the basic geometry of the parabola:

Geometric DefinitionConstruction

Reflection Property

All Parabolas are Similar and on the next page:

Conic Section

Geometric DefinitionDefinition: A parabola is the set of points in the plane that

are equidistant from a point (the focus) and a line (the

directrix.)

The following exercise should help convince you that this

definition yields the parabolas you are familiar with.

- .change? How does the directrix (vertical line) change? Find the coordinates of thefocus F and the equation of the directrix (vertical line) in terms of a.

9 - Use the slider to change a (a = 2 for example), delete the points and theparabola and repeat the steps to plots points and then the parabola.

9- Exercise: Sketch a parabola on paper. Draw a line through the focus F that isperpendicular to the axis of the parabola. This line intersect the parabola at twopoints M and N. Show that distance d(M,N) is equal to twice the distance from F tothe directrix

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 4/14

Exercise: Given a focus at (0,1) and a directrix y=-1, find the

equation of the parabola. How to do it: draw a figure showing

a generic point P on the parabola, with coordinates (x,y).Calculate its distance to the focus, its distance to the

directrix, set those equal, and simplify. Or, for a more general

result, do this exercise for a focus at (0,f) and a directrix y=-f.

↑ 

Construction

Given the focus (F) and directrix (d), here is a method toconstruct any number of points on the parabola: choose a

 point T on d. Construct the perpendicular bisector of TF.

Construct the perpendicular to d through T. The intersectionof these two lines (P) is a point on the parabola. (Make sure

you understand why.)

Exercise: With the help of dynamic geometry software, construct P as outlined above,

then trace P as T moves, or create its locus, which is the parabola. [The figure above wascreated in Cabri. You can drag F or T. To replay the construction step by step, double-

click it and use the toolbar that appears at the bottom.]

Low-tech alternative / prequel: Use this focus-directrix graph paper to construct points

on a parabola by hand. Select a line to be the directrix, and use the graph paper lines tofind points equidistant from it and the vertex.Extra Challenges (using interactive geometry software): 

• Find other constructions of the parabola, given focus and directrix. (Downloadone solution in Cabri. If you send me your Sketchpad constructions, I'll post them here and credit

you.)

• Construct a parabola given its axis of symmetry, its vertex, and one other 

 point on it.

↑ 

Reflection Property 

A light ray originating at the focus will be reflected on the parabola and continue in adirection parallel to the axis of symmetry. Likewise, a light ray coming in parallel to the

axis of symmetry will be reflected to hit the focus.

That this works is readily proved using the above construction, if you assume a basic factfrom optics: the angle of incidence equals the angle of reflection. The key to the proof is

realizing that MP must be tangent to the parabola. Indeed, if it intersected it again at a

 point P', that point would be equidistant from F and T, but it would necessarily be further 

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 5/14

from or closer to d , and thus could not be on the parabola -- a contradiction. So P' cannot

exist, and MP is a tangent.

Exercise: Prove the reflection property of the parabola, assuming that the angles of incidence and reflection are determined with respect to the tangent to the parabola at the

 point of incidence.

This property is of course the basis of many applications (headlights, flashlights, satellite

dishes, radar...) For example, here is a diagram of how this works in a reflector telescope:

The primary mirror is parabolic, reflecting the parallel rays to the focus. The secondary

(flat) mirror redirects this towards the eyepiece.

↑ 

 All Parabolas are Similar 

Like squares and circles, unlike rectangles and ellipses, all parabolas are similar. They

cannot be "pointier" or "wider". They all have exactly the same shape, which appears

"pointier" from afar, and "wider" when looked at in the neighborhood of the vertex.

Unfortunately, many of us have misled many students by implying otherwise: we oftenclaim that changing the value of a in the formula y=ax2 changes the shape of the parabola.

In fact, many teachers believe this to be true. Here are three types of arguments to show it

is a misunderstanding.

Algebraic Argument:

y=ax2

ay=a2x2

ay=(ax)2

In other words, in the equation y=x2, both x and y have been multiplied

 by the same number a. The parabola is scaled with no distortion.

Geometric Argument:

Since the directrix is infinite, moving the focus has no effect on the parabola's shape. It is merely zooming in or out on one shape.

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 6/14

Visual Argument:

Same equation, apparently different shapes:

(Dan Bennett suggests a dramatic illustration of this: make a transparency of afigure like the one above. Project it. Use another transparency to trace a piece of 

the projection, like the one below. Compare the two transparencies, which seem

to have very different shapes, but clearly must represent the same equation.)

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 7/14

Different equations, identical shape:

In fact, you can see for yourself: in the Cabri applet below, drag the axes' unit

(initially "0.2") left or right on the x-axis, and watch the "a" in the equation

change while the parabola's shape remains absolutely constant

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 8/14

Assignment 6:

"Parabola Construction"

by

Margo Gonterman

A parabola is the set of all points that are the same distance

from a line, called the directrix, and a point, called the focus.

Parabola Construction

- Construct a line that will be the directrix

- Construct a point not on the line to be the focus

- Construct a free point P on the directrix- Construct a line through P that is perpendicular to the directrix

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 9/14

- Connect P and the focus.

- Construct the perpendicular bisector of the line connecting P and

the focus through the midpoint M.

- Mark the intersection of the perpendicular bisector and the line

 perpendicular to the directrix through P as X.

-X will trace the parabola as P moves along the directrix

- Line XM is the line tangent to the parabola

Therefore:

PM=FM

Angle PMX=Angle FMX since they are both right angles

- By SAS (side-angle-side), the two triangle are congruent.

Specifically, segment PX=segment FX

- PX is the distance between X and the directrix.

- FX is the distance between X and the focus.

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 10/14

Therefore the point X is equidistant from the focus and the

directrix.

Proof that Construction Satisfies Definition 

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 11/14

- Consider triangle PMX and triangle FMX.

- Segment XM is common to both triangles.

- Line XM is the perpendicular bisector of line PF.

Therefore:

PM=FM

Angle PMX=Angle FMX since they are both right angles

- By SAS (side-angle-side), the two triangle are congruent.

Specifically, segment PX=segment FX

- PX is the distance between X and the directrix.

- FX is the distance between X and the focus.

Therefore the point X is equidistant from the focus and the

directrix

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 12/14

Here is the problem: Given a point on the curve A, and the slope of the tangent at thatpoint, (AC) and a second point on the curve B, construct (in the classical sense)additional points on the parabola... C was place above B by chance, and can be anywherealong the tangent. I have placed the problem on a coordinate grid to present it as afunction of x, but the actual coordinates of the points have no influence on theconstruction, although it is assumed that you know the direction of the axis of symmetry(in this case, vertical).

The calculus student in you might want to attack this analytically, but time for that later.Let me show you the Geometric method of Archimedes.We begin by constructing a vertical line through B, and selecting a point D, somewhere

along this line . Through this point draw another line parallel to the tangent and a secondthrough point A. Finally draw a secant AB of the parabola.

And then the final act. Mark the point where the parallel to the tangent intersects thesecant AB. From this point, extend a vertical line to find the point where it intersects AD.

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 13/14

This final point P is on the parabola AB with a tangent of AC at A, and it will be for whatever point D you picked originally.

With straight edge and compass, you would have to pick a new point D, then recreate

another parallel to the tangent, find another intersection at E, and then vertically transfer 

that up to the line AD for each new point. But with Geogebra, you can construct, and then

 just move D and watch P trace out the parabola.

So NOW let’s do a little calculus.If the original points are at (0,0) [why not] and (p,q) and the slope of the tangent is m, then

we need to find A, and B (C=0 by a clever choice of coordinates) for the parabola y= Ax2 +

Bx. We also know that at x= 0, dy/dx = m so 2Ax +B = m so B must be the slope m. Now

we just need to fill in y= Ax2 + mx and passing through (p,q). This gives us q = Ap2+mp

and we can solve for A = (q-mp)/p2.

With my selected easy values of m=1 and (p,q) = (4,1) we see that y= -3x2 /16 + x .

Two more nice problems for Calculus students that point out things that are easy not to

notice in the rush to memorize rules and such..

8/3/2019 Construct a Parabola

http://slidepdf.com/reader/full/construct-a-parabola 14/14

PROVE each:

1) If you draw to tangents to a parabolic function, the x-coordinates of their intersection is

the arithmetic average of the x-coordinates of the two points of tangency.

2) If you draw the tangents to any parabola at the endpoints of the latus-rectum, they will

always be perpendicular.