The Conics

By: Diana Brown


Day Eight:

Hyperbola Introduction


How to Construct A Hyperbola Using Wax Paper:

Start out with a piece of uncrumbled wax paper about the size of half a sheet of standard notebook paper (8.5 x 11).  Draw a circle in the center of the wax paper without writing off the paper.  Then draw a point anywhere outside the circle excluding on the circle.  The next step is to fold up the point onto the circle so they are touching.  When this point is aligned onto the circle, crease the paper and fold it accordingly.  Choose another part of the circle and align this with the point, creasing the wax paper.  Repeat this step several times until the point has touched the majority of the circle or until a hyperbola is visible.

How to Construct A Hyperbola Using Geometer’s Sketchpad:

On a new sketch, construct a circle with center A using the circle tool.   Using the point tool, draw a point B outside the circle (but not on it).

 


Select the point B and make a line segment between it and a point C on the circle.

 


Place a point D at the midpoint of that segment.  Construct the perpendicular bisector of line segment BC and while this line is still selected, choose Trace Line from the Display menu.

 



 

Highlight point C and click animate point from the Display menu and watch the hyperbola form!

 

 

Proof

Statement: The difference of the distances from two points (the foci) inside a hyperbola to any point on the hyperbola is constant.
Proof: The foci on our sketch are again the points A (the center of the circle) and B (the point outside the circle).  We will again use Geometer's Sketchpad to visualize our proof.  Construct a line through points A and C and place a point of intersection E where this new line intersects the perpendicular line passing through point D.  This point E traces out the hyperbola.  Construct a line segment between points E and B.  You should now have a triangle BCE with a perpendicular bisector ED.  Because ED is the perpendicular bisector, BD is congruent to DC and the angles BDE and CDE are both 90 degrees.  Of course, DE is congruent to itself, and thus we have two triangles with two congruent sides with an included congruent angle.  By SAS, the triangles are congruent and therefore EB is congruent to EC.  The radius of the circle is AC and AE = AC+EC.  Then AE-EC = AC and therefore AE-BE=AC.  So since the radius is constant, the difference AE-BE is always constant.

 

 

 

 


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