Ellipses and Hyperbolas

 

By Courtney Cody

 

 

In this assignment, I will construct the locus of points equidistant from a fixed point F and a circle.  Let F be any point in the plane other than the center of the circle. Assume F is not on the circle; it can be either inside or outside.

 

 

We’ll begin our construction with an arbitrary circle and a point not on the circle.  In particular, let the point not on the circle be point F and let it lie inside the circle.  In addition, let D be some point on the circle and the center of the circle is point C.

 

 

Since we want to construct the locus of points equidistant from a fixed point F and a circle, we will construct the segment between points D and F and find the midpoint of this segment.  Now, if we construct the perpendicular bisector of this segment, this represents all points equidistant from D and F. 

 

 

If we recall from our assumption, we let D be any point on the circle.  Hence, the perpendicular bisector of the segment between points D and F is specific to the location of D on the circle.  So if we drag D around the circle, we should expect numerous different perpendicular bisectors to form.  Let’s take a look at the many lines that result, which are pictured in red!

 

 

It appears as though all of the different perpendicular bisectors of the segment between D and F for all D on the circle trace an elliptical shape!!!  More specifically, the intersection of the line through D and C and the perpendicular bisector of segment DF is a point that traces an elliptical shape.  Even more, this is the point where the perpendicular bisector is tangent to the ellipse!!!

 

 

By definition, an ellipse is the set of all points such that the sum of the distances from each foci to a given point is constant.

 

Suppose we want to prove that our construction is in fact an ellipse.  Let the point that traces the ellipse be point P and let the midpoint of the segment between D and F (DF) be point M.  Also, if we construct the segments PF and PM then we notice that two triangles are form: triangle DMP (Purple) and triangle FMP (Yellow). 

 

 

Since these triangles share a side PM, then by the reflexive property, PM is congruent to PM.  Also, since the tangent line from the construction is the perpendicular bisector of segment DF, then angles DMP and FMP are both 90 degrees and are congruent.  This perpendicular bisector also implies that DM is congruent to FM.  Hence, by the side-angle-side congruence axiom the two triangles are congruent. 

 

By the definition of ellipse, we need to show that the sum of the distances from a point on the ellipse to the two foci is a constant.  Since the two triangles are congruent, then segment PD is congruent to FP.  Since P also lies on segment CP, then CP + PD = CD.  If we combine the equations together then the sum CP + FP is equal to CD, which is the radius of the circle and a constant, as desired.

 

Thus, our construction is in fact an ellipse and points C and F are the two foci of the ellipse.

 

 

Well… what if our original construction was based on point F lying outside of the circle.  How would this change our construction?  Using the same exact procedure to construct the locus of points when F lies outside of the circle, we obtain the following:

 

 

At the instant point F crossed to the outside of the circle, the locus of points changed from and elliptical shape to a hyperbolical shape!

 

By definition, a hyperbola is the set of all points such that the difference of the distances from each foci to a given point is constant.

 

Suppose we want to prove that our new construction when point F is outside of the circle is a hyperbola.  Similar to the proof of the ellipse, we have the following diagram:

 

 

Since the purple and yellow triangles share a side PM, then by the reflexive property, PM is congruent to PM.  Also, since the tangent line from the construction is the perpendicular bisector of segment DF, then angles DMP and FMP are both 90 degrees and are congruent.  This perpendicular bisector also implies that DM is congruent to FM.  Hence, by the side-angle-side congruence axiom the two triangles are congruent. 

 

By the definition of hyperbola, we need to show that the difference of the distances from each foci to a given point is constant.  Since the two triangles are congruent, then segment PD is congruent to FP.  Since P also lies on segment CP, then PD - CP = CD.  If we combine the equations together then the sum FP - CP is equal to CD, which is the radius of the circle and a constant, as desired.

 

Thus, our construction when point F lies outside of the circle is in fact a hyperbola and points C and F are the two foci of the hyperbola.

 


Return