## Exploring Analyic Geometry with Mathematica®

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General Conics

In previous chapters we have examined specific forms of an equation of the second degree resulting in a detailed understanding of circles, parabolas, ellipses and hyperbolas. In this chapter we will study the general second-degree equation itself resulting in a more complete understanding of the equation.

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`

In this section we will present a method for converting a general quadratic equation of the form to a conic curve in a standard form. The method involves examining the coefficients of the equation and applying algebraic operations to the equation which successively simplify the equation until a standard form can be recognized by inspection. The general approach involves the following steps:

• If the quadratic equation is one of several special forms, then the standard form of the curve can be determined by inspection. The following curves have a quadratic form that can be directly recognized: (1) a single point, (2) a single line, (3) two lines (parallel, coincident or intersecting), (4) a circle, parabola, ellipse or hyperbola in standard position and (5) several forms with no real locus (imaginary).

• If the quadratic equation has first-degree terms (D≠0 or E≠0), translate the equation to a coordinate system that eliminates the x or y terms. Once the curve is identified, translate the standard curve back to the original position.

• If there is an xy cross-product term in the quadratic equation (B≠0), eliminate it by applying an appropriate rotation. After the standard curve is identified, rotate it back to the original position.

The following subsections describe each of these reduction steps in more detail.

Linear Polynomial

Form: Q≡Dx+Ey+F=0, D and E not both zero.

If the first three coefficients of Q are equal to zero, and coefficients D and E are not both zero, then the equation Q represents a single straight line Dx+Ey+F=0.

Example: Show that Descarta2D will detect a quadratic equation as a line if the first three coefficients are zero. Use the line x-2y+4=0 as an example.

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Pair of Vertical Lines

Form: , A≠0.

If Q takes the form and A≠0 then Q can be factored into two linear terms using the quadratic formula. This yields the two equations

If the discriminant of this equation, , is less than zero, then there are no real points in the locus represented by Q. Otherwise, Q represents a pair of vertical lines whose equations are

The two lines are coincident if d=0.

Example: Show that the equation represents two vertical lines.

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Pair of Horizontal Lines

Form: , C≠0.

If Q takes the form and C≠0 then Q can be factored into two linear terms using the quadratic formula. This yields the two equations

If the discriminant of this equation, , is less than zero, then there are no real points in the locus represented by Q. Otherwise, Q represents a pair of horizontal lines whose equations are

The two lines are coincident if d=0.

Example: Show that the equation represents two horizontal lines.

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Intersecting Lines (or a Single Point)

Form: , A≠0 and C≠0.

If Q consists of and terms only its locus is either a single point or a pair of intersecting lines. If AC>0 then the locus is the single point (0,0). If A<0 and C>0, then the equation factors into the two linear terms

If A>0 and C<0, then the equation factors into the two linear terms given by

Both of these equations represent a pair of lines that intersect at the origin.

Example: Show that the equation represents a pair of intersecting lines.

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Circle

Form: , A=C, A≠0, C≠0, F≠0.

When the coefficients of the and terms of Q are equal and none of the coefficients A, C, or F are equal to zero, the equation has no locus if F>0; otherwise, when F<0, the locus is a circle centered at the origin with radius .

Example: Show that the equation is the equation of a circle.

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Parabola (Horizontal Axis)

Form: , C≠0 and D≠0.

When Q has a term and an x term, and the and xy terms are missing, Q represents a parabola whose axis is horizontal. The vertex, (h,k), and the focal length, f, may be determined by completing the square and forming the equation

where

which is clearly a parabola. The parabola will open to the right if f is positive and it will open to the left if f is negative.

Example: Find and plot the parabola whose equation is .

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Sketch2D[crv,CurveLength2D->60]

Parabola (Vertical Axis)

Form: , A≠0 and E≠0.

When Q has an term and a y term, and the and xy terms are missing, Q represents a parabola whose axis is vertical. The vertex, (h,k), and the focal length, f, may be determined by completing the square and forming the equation

where

which is clearly a parabola. The parabola will open upward if f is positive and it will open downward if f is negative.

Example: Find the parabola whose equation is .

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Central Conic (Ellipse or Hyperbola)

Form: , A≠0, C≠0, F≠0, and A≠C.

If Q has non-zero coefficients on the , , and constant terms, A≠C, and all the other coefficients are zero, then Q can be written in the form

This equation represents an ellipse, a hyperbola or no real locus depending of the values of -F/A and -F/C. The real loci (ellipses and hyperbolas) are centered at the origin (0,0) and have sizes and orientations as shown in the following table:

 Condition Locus a b θ no locus - - - hyperbola 0 hyperbola ellipse 0 ellipse

Example: Find and plot the conic curve whose equation is .

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Sketch2D[crv]

Remove the First-Degree Terms

Form: , A≠0, C≠0, D or E non-zero.

If both the and terms are present in Q along with at least one of the x or y terms, then Q can be simplified by introducing a change in variables. Specifically, if the substitutions

are made in Q a new equation

will result where

 A' = C' = F' =

Q' is now in a form that can be recognized by inspection. The change in variables translates the origin of the conic. To restore it to its original position we apply the inverse translation to the standard form of the conic.

Example: Find the conic whose equation is . Plot the conic.

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Sketch2D[crv]

Eliminate the xy Term

Form: , B≠0.

All quadratic equations with a non-zero xy term coefficient are standard conics in a rotated position. It can be shown that rotating such an equation by the angle θ, where

will produce a new quadratic equation, Q', whose x'y' coefficient B' will be zero (see exploration elimxy1.html).

The coefficient of the xy term can also be removed by making the substitutions

x'=kx+y   and   y'=ky-x

where

(see exploration elimxy2.html). These substitutions are equivalent to a rotation θ where

and a scaling by the factor

The equation for Q' resulting from the substitution is given by

where

 A' = C' = D' = Dk-E, E' = Ek+D   and F' = F

as shown in explorations elimxy2.html and elimxy3.html. Q' is then a quadratic equation without an xy term that can be recognized by the previously presented techniques. A scaling and rotation is applied to the resulting conic returning it to its original position. This approach is the one implemented in Descarta2D since no trigonometric functions are involved in the process, except for the final rotation.

Example: Find the conic curve represented by the equation

and plot the curve.

Solution: Use the Descarta2D function Loci2D[quad].

Clear[x,y];

Sketch2D[crv]

Classification of Conics [Top]

We may desire to determine the type of a conic from the general equation without computing the defining numerical parameters. This can be accomplished by examining the values of a set of invariant expressions. For simplicity of the invariant expressions we choose to write the quadratic equation in the form

where the factor 2 is inserted in the xy, x and y terms. For this form of the equation we define

 I = a+b, J = K =

and

Each of the four expressions is invariant under rotation of the coordinate axes; that is, they are equal respectively to the corresponding expressions after a rotation is performed. The invariants are useful in the classification of conics as shown in Table [conic:tbl02].

Classification of conics.

Center Point of a Conic [Top]

The center point of a central conic (a circle, ellipse or hyperbola) can be determined directly from its equation. The center point, (h,k), of has a relatively simple form given by

If then the conic is a parabola and has no center.

Example: Find the center point of .

Solution: The Descarta2D function Point2D[quad] returns the center point of a central conic.

Clear[x,y];

Conic from Point, Line and Eccentricity [Top]

Conic curves may be defined as the locus of a point that moves so that the ratio of its distance from a fixed point and from a fixed line is a constant. The fixed point is called the focus, the fixed line the directrix and the constant ratio the eccentricity. In previous chapters is has been shown that if the eccentricity, e, is a positive number less than one, then the conic curve is an ellipse, if e=1 a parabola and if e>1 the curve is a hyperbola.

Consider a focus point and a (normalized) directrix line λx+μy-ρ=0 (where ). The distance, , from a point P(x,y) on the locus to the focus F is given by

and the distance, , from point P to the directrix line is given by

By definition, the equation of the conic curve is

or

Squaring both sides and rearranging yields

 . .

This equation is of the form and is, therefore, a conic curve of the second degree. The equation reveals that if the defining directrix line is parallel to one of the coordinate axes, then B=0, since either λ or μ will be zero and the equation will have no xy term.

Example: Find the quadratic equation of the curve whose focus is the point (2,1), directrix is x-3y+3=0 and eccentricity is 2. Plot the conic curve.

Solution: The Descarta2D function Quadratic2D[point,line,e] returns a quadratic representing the equation of the conic with the given point as a focus, the line as a directrix and the given eccentricity.

ln1=Line2D[1,-3,3],2] //Simplify

Sketch2D[{pt1,ln1,Loci2D[q1]}]

Common Vertex Equation [Top]

All of the proper conics (circles, ellipses, hyperbolas and parabolas) can be represented by an equation involving the vertex of the conic. The expression 2p in the equation of the parabola is the length of the chord of the parabola perpendicular to the x-axis through the focus and represents a measure of the width of the parabola. The expression 2p is called the parameter of the parabola. This definition can be generalized to the other conics: the parameter of a conic is defined as the length of the chord perpendicular to the principal axis through the focus. The length of this chord for each conic is shown in Table [conic:tbl01] and is quite easy to determine from the standard form of each conic.

Parameter of a conic.

Consider the equation of an ellipse centered at the origin in standard position

Transforming the origin to the vertex V(-a,0) yields the equation

which can be rearranged into , or, by using the semi-parameter of the ellipse, into

The relation to the vertex equation of the parabola is obvious. The term is subtracted from the term 2px to obtain the ellipse. This explains the name ellipse: it is derived from the Greek term elleipsis meaning a deficiency compared with a parabola.

Similarly, the equation of a hyperbola

referred to by its vertex can be shown to be

where is the semi-parameter of the hyperbola. Compared with the parabola , there is a term in excess of the term 2px. This explains the name hyperbola from the Greek hyperbole meaning the excess.

By introducing the eccentricity e of the conic, all of the vertex equations can be represented by the common vertex equation

where (h,k) is the vertex point of the conic, 2p is the parameter of the conic and e is the eccentricity. The vertex equation also includes the case of a circle by using e=0 as the eccentricity.

Example: Plot the four curves represented by the vertex equation

for the eccentricities e={0,3/4,1,3/2}.

Solution: The Descarta2D function Loci2D[point,len,e,θ] constructs a conic (circle, ellipse, hyperbola or parabola) from the vertex point, focal chord length, eccentricity and rotation angle.

con1=Map[Loci2D[Point2D[{1,0}],1,#,0]&,{0,3/4,1,3/2}]

Sketch2D[con1, PlotRange->{{1/2,5},{-2,2}}]

Descarta2D Hint: The Descarta2D function Quadratic2D[point,len,e,θ] returns a quadratic given the vertex point, focal chord length, eccentricity and rotation angle.

Conic Intersections [Top]

Intersecting two curves is most easily accomplished if we can obtain parametric equations for one of them and an implicit equation for the other. Specifically, suppose that the first curve has parametric equations x=x(t) and y=y(t) and the second curve has an implicit equation f(x,y)=0. By substitution these two curves intersect at values of t satisfying f(x(t),y(t))=0. Once the values for t are known they can be substituted into the parametric equations to find the (x,y) coordinates of the intersection points.

As a specific application of this technique, suppose we wish to find the intersection points of a line px+qy+r=0 and a conic curve . We can take either x or y as the parameter of the equation px+qy+r=0; suppose we select x (assuming q≠0), yielding the parametric equations

Substituting these values into the equation for the conic curve yields a quadratic equation in the variable x given by

which is easy to solve using the quadratic formula. Once the two values for x are known, the corresponding values for y can be determined using the parametric equations of the line. So, in the general case, a line and a conic will intersect in two points, the points being real and distinct, real and coincident (the line being tangent to the conic) or imaginary (the line does not intersect the conic).

Now consider the case of two intersecting conic curves whose equations are given by

 = 0 = 0

Depending on the values of the coefficients it may or may not be easy to express one of the equations with a pair of parametric equations; therefore, we look for alternative techniques for finding the points of intersection. The brute force approach to the problem is to simply regard it as a problem of solving two non-linear equations in two unknowns. Mathematica can solve such systems of equations both numerically and symbolically, and this is, in fact, the method implemented in Descarta2D.

Alternatively, the method of pencils can be used. Suppose we have two curves f(x,y) and g(x,y). For any given value of λ, we can form the equation f(x,y)+λ  g(x,y)=0 which obviously passes through all the points of intersection of the original two curves. As λ varies, an entire family of curves, called a pencil, will be produced. By selecting an appropriate value for λ we can hope to produce an equation f(x,y)+λ  g(x,y)=0 that is particularly simple. We can then intersect the simpler curve with one of the original curves. This approach works well for conic curves because there always exists a value for λ such that f(x,y)+λ  g(x,y)=0 represents two straight lines. Intersecting these two lines with either of the original conics produces the four intersection points (which may be real and distinct, real and coincident or imaginary). So there may be up to four points of intersection between two conic curves. Since there exist three pairs of lines passing through four points, there are three values for λ that represent two lines in the equation of the pencil. Finding the three values for λ involves solving a cubic equation, which appears to be easier than solving two non-linear equations in two unknowns. (Since solving two non-linear equations in two unknowns is equivalent to solving a fourth-degree equation, and solving a fourth-degree equation reduces to solving a cubic equation, the two techniques are mathematically similar in complexity.)

Example: Find the points of intersection of the line x-2y+2=0 with (a) the circle and (b) the ellipse .

Solution: The Descarta2D function Points2D[curve,curve] returns a list of points that are the intersection of the two curves.

l1=Line2D[1,-2,2];
c1=Circle2D[{0,0},2];
e1=Ellipse2D[{0,0},3,1,0];
pts={Points2D[l1,e1],Points2D[c1,e1]}//N

Sketch2D[{l1,c1,e1,pts}]

Explorations [Top]

Eliminate Cross-Term by Rotation. [elimxy1.html]

Show that rotating a quadratic through an angle θ given by

will cause the xy term to vanish.

Eliminate Cross-Term by Change in Variables. [elimxy2.html]

Show that applying the change in variables x'=kx+y and y'=ky-x, where

to the equation will cause the xy term to vanish and a new quadratic with the following coefficients will be formed:

 a' = b' = 0, c' = d' = dk-e, e' = ek+d   and f' = f.

Eliminate Cross-Term by Change in Variables. [elimxy3.html]

Show that applying the change in variables x'=kx+y and y'=ky-x, where

to the equation is equivalent to rotating the quadratic by an angle θ given by

and scaling the quadratic by a scale factor

Eliminate Linear Terms. [elimlin.html]

Show that applying the change in variables

whose linear terms have vanished.

Show that applying the change in variables

to the quadratic causes the linear terms to vanish, implying that the center of the conic is

Polar Equation of a Conic [polarcon.html]

Let the focus F of a conic be at the pole of a polar coordinate system and the directrix D be perpendicular to the polar axis at a distance ρ to the left of the pole as shown in the figure. Show that the polar equation of the conic is

where e is the eccentricity of the conic.