Hyperbola from Focus and Directrix

hypfd.html

Exploration

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

where

and .

Approach

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

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 hyperbola is the angle the line perpendicular to L makes with the +x-axis (in Mathematica ArcTan[p,q] is ArcTan[q/p], the first form takes into account the quadrantof 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.

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).

Solve[e==Sqrt[a^2+b^2]/a,b]

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/(e^2-1),
    b->a*Sqrt[e^2-1],
    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 a hyperbola from its focus, directrix and eccentricity.

focus1=Point2D[{1/2,1}];
directrix1=Line2D[5,8,-15];
eccentricity1=5/4;
hyperbola1=Hyperbola2D[focus1,directrix1,eccentricity1]

Sketch2D[{focus1,directrix1,hyperbola1},
   CurveLength2D->5]

Graphics saved as "hypfd01.eps".