## Exploring Analyic Geometry with |
|||||

Home | Contents | Commands | Packages | Explorations | Reference |

Tour | Lines | Circles | Conics | Analysis | Tangents |

Areas Related to Hyperbolas

hyparea.html

Exploration

Graphics saved as "area07.eps".

Graphics saved as "area06.eps".

Referring to the figures, use calculus to verify that the areas between two parameters and of a segment and a sector of a hyperbola are given by

where a and b are the lengths of the semi-transverse and semi-conjugate axes, respectively, and e is the eccentricity of the hyperbola (assuming the parameterization Descarta2D uses for a hyperbola).

Approach

Find the coordinates of and , the coordinates of the ends of the infinitesimal rectangle. Integrate from to to find the area of the segment. Find the area of the from its vertex points. Subtract the area of the segment from the area of the triangle to find the area of the sector.

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 x-coordinate of a point on the hyperbola (found by solving for x) in terms of the y-coordinate.

Clear[x,y,a,b];

X1=a*Sqrt[1+y^2/b^2]

The x-coordinate of a point on a line between and This is found by intersecting a horizontal line through the point on the hyperbola with the line between and .

Clear[x1,y1,x2,y2];

X2=XCoordinate2D[

Point2D[Line2D[{x1,y1},{x2,y2}],

Line2D[Point2D[x,y],0]]] //FullSimplify

L is the length of the horizontal line segment between the hyperbola and the line through and .

L=X2-X1 //FullSimplify

Find the indefinite integral L Δy, which represents an infinitesimal rectangular area.

I1=Integrate[L,y] //FullSimplify

Find the area of the hyperbolic segment between the chord and the hyperbola by evaluating the integral at the vertical limits. Simplify.

A1=(I1 /. y->y2)-(I1 /. y->y1) //FullSimplify

A2=A1 //. {

a*Sqrt[1+y1^2/b^2]->x1,

a*Sqrt[1+y2^2/b^2]->x2}

Create the hyperbola.

H1=Hyperbola2D[{0,0},a,b,0];

Find the coordinates of a point at a general parameter t on a hyperbola.

Clear[s];

P=H1[t] /. ArcCosh[Sqrt[a^2+b^2]/a]->s

Simplify.

Clear[t1,t2];

A3=A2 //. {

x1->(P[[1]] /. t->t1),

x2->(P[[1]] /. t->t2),

y1->(P[[2]] /. t->t1),

y2->(P[[2]] /. t->t2)} //FullSimplify

Clear[E1];

A4=A3 /.

ArcSinh[Sinh[E1_]]->E1 //FullSimplify

AreaSegment=A4 /.

Sinh[s(t1-t2)]->-Sinh[s(t2-t1)]

Find the area of the triangle .

AreaTriangle=Area2D[

Triangle2D[{0,0},{x1,y1},{x2,y2}]] /. {

x1->(P[[1]] /. t->t1),

x2->(P[[1]] /. t->t2),

y1->(P[[2]] /. t->t1),

y2->(P[[2]] /. t->t2)} //FullSimplify

The area of the sector is the difference.

AreaSector=AreaTriangle-AreaSegment //Simplify

www.Descarta2D.com