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Ellipse from Focus and Directrix

ellfd.html

Exploration

Show that the ellipse with focus , directrix line L≡ p x+q y+r=0 and eccentricity, 0<e<1, is defined by the constants

where

Approach

Apply the definition of an ellipse to the supplied focus and directrix for a general point (x,y) and show that the derived locus is an ellipse.

Initialize

To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.

This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.

<<Descarta2D`

Solution

The rotation angle of the ellipse is the angle the line perpendicular to L makes with the +x-axis (in Mathematica ArcTag[p,q] is ArcTan[q/p], the first form takes into account the quadrant of the point (p,q)).

Clear[p,q,r];

L=Line2D[p,q,r];

theta=Angle2D[Line2D[0,1,0],Line2D[Point2D[0,0],L]];

theta //Simplify

Now we must show that the lengths a and b are given by the formulas. In standard position the distance from the focus of an ellipse to its directrix is given by d=a/e-a e. Solving for a gives the following result.

Clear[d,a,e];

Solve[d==a/e-a*e,a] //Simplify

Also, the eccentricity is given by and solving for b gives (take the positive result).

FullSimplify[Solve[e==Sqrt[a^2-b^2]/a,b],Assumptions->{a>0,e>0,e<1}]

The eccentricity is the ratio of the distance from a general point to the focus to the distance to the directrix.

Clear[x1,y1,x,y];

F=Point2D[x1,y1];

P=Point2D[x,y];

{dF=Distance2D[P,F],

dL=Distance2D[P,L]}

Form the equation for the eccentricity squared.

eq1=e^2*dL^2-dF^2 //Expand //Together

Find the coordinates of the center of the quadratic.

{h1,k1}=

Coordinates2D[

Point2D[

Q1=Quadratic2D[eq1,{x,y}]//Simplify]] //Simplify

Find the coordinates of the center using the formula provided.

Clear[D1];

{h2,k2}={x1+p*a*e*D1/d, y1+q*a*e*D1/d} //.

{a->d*e/(1-e^2),

b->a*Sqrt[1-e^2],

d->Sqrt[(p*x1+q*y1+r)^2/(p^2+q^2)],

D1->(p*x1+q*y1+r)/(p^2+q^2)}

This shows that the center indeed has the same coordinates as the point from the formula.

{h1-h2,k1-k2} //Simplify

Discussion

An example showing the construction of an ellipse from its focus, directrix and eccentricity.

focus1=Point2D[{1/2,1}];

directrix1=Line2D[5,8,-20];

eccentricity1=3/4;

ellipse1=Ellipse2D[focus1,directrix1,eccentricity1]

Sketch2D[{focus1,directrix1,ellipse1},

CurveLength2D->5]

Graphics saved as "ellfd01.eps".

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