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Tangent Conics

Constraint Equations

Systems of Quadratics

Validity Conditions

Five Points

Four Points, One Tangent Line

Three Points, Two Tangent Lines

Conics by Reciprocal Polars

Explorations

The most general quadratic equation in two unknowns

has six coefficients, but since we can divide the coefficients by any non-zero constant, say F, without altering the equality obtaining

a quadratic equation only has five degrees of freedom. Thus we may specify five conditions (or constraints) on a quadratic equation. In this chapter we will investigate the construction of conic curves (circles, parabolas, ellipses and hyperbolas) that satisfy a set of five conditions, when the conditions are of two specific types: either passing through a given point, or tangent to a given line. The resulting equations are sufficiently complex that obtaining the solutions in symbolic, closed form is of no practical value, so we will illustrate the solution techniques and use the numerical capabilities of Mathematica to compute specific solutions.

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`

Constraint Equations [Top]

As mentioned in previous chapters, if the curve represented by the quadratic equation

passes through the point , then the point will satisfy the equation yielding the relationship

It has also been shown in previous chapters that the line px+qy+r=0 will be tangent to the curve represented by the quadratic equation if the coefficients satisfy the equation

2(BD-2AE)qr+2(BE-2CD)pr+2(DE-2BF)pq=0. |

Systems of Quadratics [Top]

In this section we will outline the general technique for finding quadratics that pass through the four points of intersection of two quadratic curves. These techniques will be the basis for subsequent sections wherein we will find quadratics satisfying a variety of conditions.

Two quadratics intersect in four points (four real, two real and two complex, or four complex) since each equation is of the second degree. If

≡ | ||

≡ |

represent the equations of the two quadratics, then , for any constant k, is the equation of a quadratic through the points of intersection of and . The equation Q is called a system or pencil of quadratics, and placing one additional condition on the equation for Q allows us to solve for k and find a specific quadratic in the pencil. The equation of the pencil is sometimes written as in order to allow the original quadratics, and , to be in the pencil (for k=0 and k=1, respectively).

Example: Find the quadratic that passes through the intersection of the ellipse and the hyperbola and also passes through the point (-4,0).

Solution: The equation of the quadratic pencil containing the solutions is

and this must be satisfied by (-4,0). Hence, solving for k yields k=1 and the final equation of the conic sought is (a parabola).

Sketch2D[{Quadratic2D[1,0,4,-10,0,-39],

Quadratic2D[-1,0,1,0,0,-1],

Quadratic2D[0,0,1,-2,0,-8],

Point2D[{-4,0}]},

CurveLength2D->40,

PlotRange->{{-10,15},{-6,6}}]

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Descarta2D Hint: Quadratic2D[quad,quad,k,Pencil2D] returns a quadratic parameterized by the variable k representing the pencil of quadratics passing through two quadratics.

A Degenerate Case

If Q=0 is the equation of a quadratic and L=0 is the equation of a straight line, then is the equation of a quadratic tangent to Q at the intersection points of Q=0 and L=0. We may think of as a (degenerate) quadratic (two coincident lines) intersecting Q=0 in two pairs of coincident points each.

Example: Find the quadratic tangent to at the points of intersection of Q and L≡3x-2y-1=0 and passing through the point (-1,0).

Solution: The equation is of the form

Substituting (-1,0) into this equation we get k=1/8, which yields as the equation of the conic sought

Sketch2D[{Quadratic2D[1,0,-1,0,-1,1],

Line2D[3,-2,-1],

Quadratic2D[1,-12,12,-6,12,-7],

Point2D[{-1,0}]}]

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Validity Conditions [Top]

In the remainder of this chapter we will outline techniques for finding conics that satisfy five conditions. The conditions will be of two types: either passing through a given point, or tangent to a given line. The following assumptions are made with respect to the five points and/or lines:

• no pair of points is coincident,

• no pair of lines is coincident,

• no triple of points is collinear,

• no triple of lines is concurrent,

• no triple of lines is mutually parallel,

• no more than one point is on each line, and

• no more than one line passes through each point.

Degenerate conics may exist that satisfy configurations of points and lines that violate these restrictions, but we will focus our attention on cases that produce proper conics (circles, parabolas, ellipses and hyperbolas).

Five Points [Top]

Quadratic through five points.

Given five points, , , , and , satisfying the validity conditions stated in Section [tcon:sec01], we wish to find the quadratic that passes through all five points. Consider the lines , , and passing through the points in pairs as shown in Figure [tcon:fig20]. Let be a (degenerate) quadratic (a pair of lines) passing through , , and . Similarly, let be a second quadratic passing through the same four points. The equation will then represent the pencil of quadratics parameterized by the variable k passing through the four points.

Applying Equation ([tcon:eqn01]), by substituting the coordinates of the point into the equation for Q, we can solve this linear equation for the value of k, thereby yielding the specific quadratic in the pencil of quadratics that passes through all five points. Mathematica can be used to solve for k and form the symbolic equation for Q, although the result is quite cumbersome in expanded form. A determinant can be used to represent the resulting quadratic in a more convenient and simplified form and is given by

Example: Find the quadratic passing through the five points (3,0), (3,1), (0,1), (-3,0) and (0,-1).

Solution: The function Quadratic2D[pt,pt,pt,pt,pt] returns the quadratic passing through the five points.

pts={p1=Point2D[{3,0}],p2=Point2D[{3,1}],

p3=Point2D[{0,1}],p4=Point2D[{-3,0}],

p5=Point2D[{0,-1}]};

q1=Quadratic2D[p1,p2,p3,p4,p5]

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Example: Find the conic represented by the quadratic found in the previous example. Plot the points and the conic curve.

Solution: The conic can be determined directly from the result of the previous example using the function Loci2D[quad]. Descarta2D also provides the function TangentConics2D[{pt,pt,pt,pt,pt}] that constructs a list containing the conic directly from the five points.

{Loci2D[q1], crv1=TangentConics2D[pts]} //N

Sketch2D[{pts,crv1}]

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Descarta2D Hint: TangentQuadratics2D[{pt,pt,pt,pt,pt}] constructs a list containing the single quadratic passing through five points. Except for the fact the TangentQuadratics2D checks for the validity conditions stated in Section [tcon:sec01], this function is equivalent to Quadratic2D[pt,pt,pt,pt,pt].

Four Points, One Tangent Line [Top]

In this section we will consider the construction of quadratics and conics passing through four points and tangent to a line. Two cases are distinguished: the first constructs the quadratic or conic when none of the given points lie on the tangent line; the second constructs the quadratic or conic when one of the given points does lie on the tangent line.

Points Not on a Tangent Line

Four points, one line, no points on the line.

Assume that points , , and and line as shown in Figure [tcon:fig15] satisfy the validity conditions stated in Section [tcon:sec01] and that none of the four points lie on . To find the equation of the quadratic passing through the four points and tangent to the line, we form a pencil of quadratics passing through the four points parameterized by the variable k that is given by

We now apply the condition that line is tangent to Q by using Equation ([tcon:eqn02]) resulting in a quadratic equation in the variable k. Solving this equation yields two values for k that can be substituted into the equation for Q, giving two quadratics satisfying the stated conditions.

Example: Find the quadratics passing through the points (2,1), (-2,1), (-2,-1) and (2,-1) and tangent to the line 3x+4y-12=0. Plot the conic curves associated with the quadratics.

Solution: The Descarta2D function TangentQuadratics2D[{pt,pt,pt,pt,ln}] constructs a list of quadratics passing through the four points and tangent to the line. TangentConics2D[{pt,pt,pt,pt,ln}] constructs a list of conic curves passing through the four points and tangent to the line. Both functions allow the points and line to be listed in any order.

objs={Point2D[{2,1}],Point2D[{-2,1}],

Point2D[{-2,-1}],Point2D[{2,-1}],

Line2D[3,4,-12]};

TangentQuadratics2D[objs]

crvs=TangentConics2D[objs] //N

Sketch2D[{objs,crvs}]

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One Point on Tangent Line

Four points, one line, one point on the line.

We now examine the case when one of the four points is on the tangent line. Consider the points , , and and the line as shown in Figure [tcon:fig16] satisfying the validity conditions stated in Section [tcon:sec01], where the point is on . Since the desired quadratic is tangent to at , we can consider to be a pair of coincident intersection points of the pencil of quadratics passing through the four points , and ( is counted as two coincident intersection points). We now form the pencil of quadratics parameterized by the variable k and given by The coordinates of the remaining point, , must satisfy the equation of the quadratic, and by applying Equation ([tcon:eqn01]) we generate a linear equation in the variable k that can be solved yielding the single quadratic equation satisfying the stated conditions.

Example: Find the conic curve passing through the points (2,0), (0,1), (-2,0) and (0,-1) and tangent to the line y=1.

Solution: The function TangentConics2D[{pt,pt,pt,pt,ln}] returns a list of conic curves passing through four points and tangent to a line. The points and line may be listed in any order.

crv=TangentConics2D[

objs={Point2D[{2,0}],Point2D[{0,1}],

Point2D[{-2,0}],Point2D[{0,-1}],

Line2D[0,1,-1]}]

Sketch2D[{objs,crv},PlotRange->{{-3,3},{-1.5,1.5}}]

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Descarta2D Hint: In the remaining sections of this chapter we will use the function TangentConics2D to find the tangent curves satisfying a variety of conditions. The function TangentQuadratics2D is also available in all these cases and will return a list of quadratics instead of a list of conic curves.

Three Points, Two Tangent Lines [Top]

We now consider the construction of a conic passing through three points and tangent to two lines. Three cases need to be considered: the first constructs the conic when none of the given points lie on either of the tangent lines; the second constructs the conic when one of the given points lies on one of the tangent lines; and, finally, two points lie on two of the tangent lines.

Points Not on Tangent Lines

Three points, two lines, no points on the lines.

Consider three points , and and two lines and satisfying the validity conditions stated in Section [tcon:sec01]. We also assume that none of the points are on either line as shown in Figure [tcon:fig19]. The line passing through the points of tangency between lines and and the desired conic curve can be written in the form

assuming we can guarantee that does not pass through the origin. The point is clearly not on because that would imply that the conic passes through three distinct, collinear points, which clearly violates the validity conditions. Therefore, if we translate all five of the original objects so that point is at the origin, we can guarantee that does not pass through the origin. Of course we need to perform the inverse translation on the resulting conic curves to produce the solution for the geometry in its original position.

We now proceed with , a line that does not pass through the origin. Consider the pencil of quadratics parameterized by the variable k and represented by the equation

Solving this equation for k yields

The right side of this equation is an expression in x and y involving the unknowns a and b. The expression must produce the same value for k for any point (x,y) on the desired quadratic. In particular, points , and must all produce the same value for k. Using the expression to indicate the expression f evaluated at the point , we can write the system of equations

since all of these expressions must equal k. Rewriting these equations as a system of two equations and cross-multiplying yields two quadratic equations in two unknowns, a and b,

= | ||

= |

Solving these equations for a and b yields four pairs of solutions which can be substituted into

producing four quadratics Q satisfying the stated conditions. The resulting quadratics may be translated back to the original position of the defining objects by translating the origin back to .

Example: Find the conics passing through (1,0), (0,-1), and tangent to the lines x=1 and y=-1.

Solution: The function TangentConics2D[{pt,pt,pt,ln,ln}] returns a list of conic curves passing through three points and tangent to two lines. The points and lines may be listed in any order.

objs={Point2D[{1,0}],Point2D[{0,-1}],

Point2D[{1/Sqrt[2],-1/Sqrt[2]}],

Line2D[0,1,-1],Line2D[1,0,1]};

crvs=TangentConics2D[objs] //N

Map[Print[Sketch2D[{objs,#},

PlotRange->{{-2,4},{-4,2}}]]&,

crvs];

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One Point on Tangent Line

Three points, two lines, one point on a line.

We now consider the case where one of the three points is on one of the two tangent lines. Assume that point is on line and points and are on neither line or as shown in Figure [tcon:fig18]. Also, we assume the points and lines satisfy the validity conditions stated in Section [tcon:sec01]. We form the pencil of quadratics

where , and are the lines passing through points and , and and and , respectively. We now apply the tangency condition by using Equation ([tcon:eqn02]) with Q and line to form a quadratic equation in the variable k. Solving the equation for k gives the two quadratics passing through the points and tangent to the lines.

Example: Find the conic curves passing through the points (0,1), (-3,0) and (0,-1) and tangent to the lines y=1 and x-2y-3=0.

Solution: The function TangentConics2D[{pt,pt,pt,ln,ln}] returns a list of conic curves passing through three points and tangent to two lines. The points and lines may be listed in any order.

objs={Point2D[{0,1}],Point2D[{-3,0}],Point2D[{0,-1}],

Line2D[0,1,-1],Line2D[1,-2,-3]};

crvs=TangentConics2D[objs] //N

Sketch2D[{objs,crvs}]

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Two Points on Tangent Lines

Three points, two lines, two points on the lines.

Let be a point on line , be a point on line , and a point not on either line as shown in Figure [tcon:fig17] and assume that these points and lines satisfy the validity conditions stated in Section [tcon:sec01]. We form the pencil of quadratics

where is the line passing through points and . We now apply Equation ([tcon:eqn01]) establishing the condition that must be on Q and, therefore, the coordinates must satisfy Q. We can solve this linear equation for k and determine the coefficients of the quadratic Q satisfying the stated conditions.

Example: Find the conic curve passing through the points (2,1), (-2,1) and (0,2) and tangent to the lines x-3y+1=0 and 2x+y+3=0.

Solution: The function TangentConics2D[{pt,pt,pt,ln,ln}] returns a list of conic curves passing through three points and tangent to two lines. The points and lines may be listed in any order.

objs={Point2D[{2,1}],Point2D[{-2,1}],Point2D[{0,2}],

Line2D[1,-3,1],Line2D[2,1,3]};

crvs=TangentConics2D[objs] //N

Sketch2D[{objs,crvs}]

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Notice that a ConicArc2D object is a special case of this construction where the start and end points are the points and and the apex point defines the lines and .

Conics by Reciprocal Polars [Top]

In this section we will introduce the concept of reciprocal polars, and a technique that will allow us to solve tangent conic problems involving more than two tangent lines. Proofs of all the concepts involved in these techniques are beyond the scope of this book, but applying the techniques to solve tangent conic problems is easily grasped.

Let C be a circle in the plane and P a point. The reciprocal of P with respect to C is simply the polar line of P with respect to C. Similarly, let L be a line. The reciprocal of L with respect to C is the pole point of L with respect to C. It is noteworthy that the center point of the circle has no reciprocal, and any line passing through the center of the circle has no reciprocal.

If we have any figure consisting of any number of points and straight lines, and we take the polars of those points and the poles of the lines, with respect to a circle C, we obtain another figure which is called the polar reciprocal of the former with respect to the auxiliary circle C. When a point in one figure and a line in the reciprocal figure are pole and polar with respect to the auxiliary circle, C, the point and line are said to correspond to one another.

An important theorem from the analytic geometry of conics states that taking the reciprocal of all the points of a conic Q with respect to some auxiliary circle C will produce an envelope of lines tangent to another conic Q'. Furthermore, any line tangent to Q will correspond to a point P' on Q', and any line L' tangent to Q' will correspond to a point P on Q (always using C as the auxiliary circle).

We use this theorem as follows to find conics tangent to three, four or five lines and passing through a corresponding number of points so the total number of conditions equals five. First we apply an arbitrary translation to the objects to insure that none of the points lie at the origin and that none of the lines pass through the origin. We now take the reciprocal of the points or lines with respect to a unit circle centered at the origin, thereby producing a new figure of corresponding lines and points. Note that if there are three or more lines in the original figure, there will be two or fewer lines in its reciprocal.

We now apply the techniques developed in the previous sections of this chapter to find the quadratics(s) satisfying the conditions imposed by the elements in the reciprocal figure. Finally, we find the reciprocal of the resulting quadratic with respect to the auxiliary circle yielding the sought-after quadratics in the original figure. If the equation of the quadratic in the reciprocal figure is

then its equation in the original figure is given by

. | . | |

. | . |

The validity of this relationship is demonstrated in the exploration recquad.html. The relationship between Q and Q' is only valid when the auxiliary circle is a unit circle at the origin ().

Two Points, Three Tangent Lines

Example: Find the conic curves passing through the points (3,-1) and (1,0) and tangent to the lines 4x-y-3=0, x+2y-3=0 and y=-2.

Solution: The function TangentConics2D[{pt,pt,ln,ln,ln}] returns a list of conic curves passing through two points and tangent to three lines. The points and lines may be listed in any order. If neither point is on any of the lines, there are at most four real conic curves; if one of the points is on one of the lines, then there are at most two real conic curves; if two of the points are on the tangent lines, then there is at most one real conic curve.

objs={Point2D[{3,-1}],Point2D[{1,0}],Line2D[4,-1,-3],

Line2D[1,2,-3],Line2D[0,1,2]};

crvs=TangentConics2D[objs] //N

Sketch2D[{objs,crvs},

CurveLength2D->20,

PlotRange->{{-1,8},{-4,2}}]

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One Point, Four Tangent Lines

Example: Find the conic curves passing through the point (-1,1) and tangent to the lines 4x-y-3=0, x+2y-3=0, x=-3 and y=-2.

Solution: The function TangentConics2D[{pt,ln,ln,ln,ln}] returns a list of conic curves passing through a point and tangent to four lines. The points and lines may be listed in any order. If the point is not on any of the lines, there will be at most two real conic curves; if the point is on one of the lines, then there will be at most one real conic curve.

objs={Point2D[{-1,1}],Line2D[4,-1,-3],

Line2D[1,2,-3],Line2D[1,0,3],Line2D[0,1,2]};

crvs=TangentConics2D[objs] //N

Sketch2D[{crvs,objs},PlotRange->{{-4,2},{-3,3}}]

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Five Tangent Lines

Example: Find the conic curve tangent to the five lines x-2y+3=0, x=3, 2x-3y-2=0, y=2 and x=-2. Plot the lines and the conic curve.

Solution: The function TangentConics2D[{ln,ln,ln,ln,ln}] returns a list of at most one conic curve tangent to five lines.

objs={Line2D[1,-2,3],Line2D[1,0,-3],

Line2D[2,-3,-2],Line2D[0,1,-2],Line2D[1,0,2]};

crvs=TangentConics2D[objs] //N

Sketch2D[{objs,crvs},PlotRange->{{-4,4},{-2,3}}]

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Explorations [Top]

Reciprocals of Points and Lines. [recptln.html]

Show that the polar reciprocal of in the auxiliary conic is the point , assuming that the line does not pass through the origin. Also, show that the line x+y-1=0 is the polar reciprocal of the point (x,y) with respect to C.

Reciprocal of a Circle. [reccir.html]

Given a circle show that its polar reciprocal in the auxiliary conic is given by the quadratic

Furthermore, show that Q is an ellipse, if the origin (0,0) is inside C; a parabola, if the origin is on C; and a hyperbola, if the origin is outside C.

Reciprocal of a Quadratic. [recquad.html]

Given a general quadratic show that the reciprocal of Q is the quadratic

. | . | |

. | . |

when the auxiliary conic is .

Parabolas Through Four Points. [pb4pts.html]

Describe a method for finding the two parabolas passing through four points. Show that the technique produces the correct results for the points (2,1), (-1,1), (-2,-1) and (4,-3) by plotting the parabolas and the four points.

Equilateral Hyperbolas. [hyp4pts.html]

Describe a method for finding the equilateral hyperbola(s) passing through four points. Show that the technique produces the correct results for the points (2,1), (-1,1), (-2,-1) and (4,-3) by plotting the hyperbola(s) and the four points.

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