L24-conicSections.mws

Calculus II

Lesson 24: Conic Sections

Geometric Definition of Conic Sections

We start with the geometric definition of conic sections. Later in this lesson, we'll plot conic sections in the plane using their analytic representations.

For each picture below, the conic section is the intersection of the cone with the plane. Grab the picture with the mouse and rotate it to see the intersection from many points of view.

> restart: with(student):with(plottools): with(plots): 

Warning, the name changecoords has been redefined

Circle

> up := cone([.3,0,-1],1,2,color=blue):
down := cone([.3,0,-1],1,-2,color=blue):
circlePlane := plot3d(0, x=-3..1.5, y=-1..1, color=green, style=patchnogrid):
display([circlePlane,up,down], scaling=constrained, orientation=[140,70], style=wireframe);

[Maple Plot]

Ellipse

> ellipsePlane := plot3d(.75*x, x=-3..1.5, y=-1..1, color=green, style=patchnogrid):
display([ellipsePlane, up,down], scaling=constrained, orientation=[140,70], style=wireframe);

[Maple Plot]

Parabola

> parabolaPlane := plot3d(4*x, x=-.75..(.25), y=-1..1, color=green, style=patchnogrid):
display([parabolaPlane, up,down], scaling=constrained, orientation=[140,70], style=wireframe);

[Maple Plot]

Hyperbola

> HyperbolaPlane := implicitplot3d(x=0, x=-.6..(.2), y=-1..1, z=-3..1,color=green, style=patchnogrid):
display([HyperbolaPlane, up,down], scaling=constrained, orientation=[140,70], style=wireframe);

[Maple Plot]

Analytic Representation of Conic Sections

Example 1

Consider the following conics. First try to identify them. Then confirm your guess by plotting.

a) x^2+y^2 = 9

b) x*y = -4

c) 4*x^2+y^2 = 16

d) x^2-4*y^2 = 36

e) y^2 = 4*x

f) x^2+3*y^2-1 = 0

g) x^2+3*x*y+y^2 = 16

h) x^2+4*y = 4

In each case, state the conic obtained along with 'specifics'.

Circle specifics = center and radius

Ellsipse specifics = center, foci, vertices

Parabola specifics = vertex, directirx, focus

Hyperbola specifics = center, vertices, foci, asymptotes

a) Graph is a circle.

Center is at (0,0).

Radius = 3.

> implicitplot( x^2 + y^2 = 9, x = -10..10, y = -10..10, scaling=constrained, grid=[40,40]);

[Maple Plot]

b) Graph is a rotated hyperbola.

> implicitplot(x*y = -4, x = -10..10, y = -10..10, grid=[40,40]);

[Maple Plot]

c) Graph is an ellsipse.

Center (0,0)

Foci: (0, +- sqrt(12) )

major axis vertices: (0,+- 4)

minor axis vertices: (+-2, 0)

> implicitplot(4* x^2 + y^2 = 16, x = -10..10, y = -10..10, grid=[50,50]);

[Maple Plot]

d) Graph is a hyperbola.

Center(0,0)

Vertices: (+- 6, 0)

Foci( +- 3*sqrt(5) , 0)

asymptotes: y = +- 1/2*x

> implicitplot(x^2 - 4* y^2 = 36, x= -10..10, y = -10..10, grid=[40,40]);

[Maple Plot]

e) Graph is a parabola.

Vertex: (0,0)

Focus: (1,0)

Directrix: x = -1

> implicitplot(y^2 = 4*x, x = -10..10, y = -10..10, grid=[40,40]);

[Maple Plot]

f) The graph is an ellispe.

Center: (0,0)

Foci: ( +- sqrt(2/3) , 0)

Major axis vertices: (+- 1,0)

Minor axis vertices: (0, +- sqrt(1/3)

> implicitplot(x^2 + 3* y^2 - 1 = 0, x = -1..1, y = -1..1, grid=[40,40]);

[Maple Plot]

g) Graph is a rotated hyperbola.

> implicitplot(x^2 + 3*x * y + y^2 = 16, x = -10..10, y = -10..10, grid=[40,40]);

[Maple Plot]

h) Graph is a parabola

Vertex: (0,1)

Focus: (0,0)

Directrix: y = 2

> implicitplot(x^2 + 4*y = 4, x = -3..3, y = -3..3, grid=[40,40]);

[Maple Plot]

> completesquare(x^2 + y^2 - 2*x + 2*y + 2 = 0,x);

(x-1)^2+1+y^2+2*y = 0

> completesquare(%,y);

(y+1)^2+(x-1)^2 = 0

Solution is one point: (1,-1).

>

Equations of Conics in Standard Form

Example 2

Write the equation of the circle that goes through (0,0), (-4,0), and (0,6).

Determine the center and radius of the circle.

> solve({f=0, 16 - 4*d + f =0, 36 + 6*e + f = 0},{d,e,f});

{d = 4, e = -6, f = 0}

> completesquare(x^2 + y^2 + 4*x - 6*y = 0, x);

(x+2)^2-4+y^2-6*y = 0

Thus, center is: (-2,3) and radius is sqrt(13) .

> completesquare(2*x^2 + 2*y^2 - 5*x -9*y + 11 = 0,x);

2*(x-5/4)^2+63/8+2*y^2-9*y = 0

Standard form: (x-5/4)^2+(y-9/4)^2 = 9/8 .

center: 5/4, 9/4

radius: sqrt(9/8) .

a) we use: (y-k)^2 = 4*p*(x-h) .

p = |3 - (-5)| = 8.

h = 3, k = -2.

Equation is: (y+2)^2 = 32*(x-3) .

vertex is (3,-2).

directrix is: x = -5.

focus is: (11,-2).

> A:= implicitplot(y^2 = 20 * x, x = -10..10, y = -10..10):

> C:= textplot([-7,7,'directrix'], color = blue):

> E:= textplot([5.5,-2,'focus'], color = blue):

> B:= plot([-5,t, t = -10..10], color = blue):

> DD:= pointplot([11,-2]):

> display({A,B,C,E,DD});

[Maple Plot]

> completesquare(%,y);

[Maple Plot]

Example 3

Write the equation of the ellipse in standard form and determine its center, vertices, foci, and covertices.

Plot the ellipse with Foci.

4*x^2+5*y^2-24*x-10*y+17 = 0

> completesquare(%,x);

[Maple Plot]

Standard form: (x-3)^2/6+(y-1)^2/(24/5) = 1 .

Center: (3,1)

a^2 = 6 and b^2 = 24/5 . Use a^2-b^2 = c^2 to get: c^2 = 6/5

Vertices: (3 +- sqrt(6) ,1)

Covertices: (3, 1 +- sqrt(24/5) )

Foci: ( 3 +- sqrt(6/5) ,1).

> A:= implicitplot(4 * x^2 + 5*y^2 - 24*x - 10*y + 17 = 0, x =-1..6 ,y = -3..4):

> C:= pointplot([3 -1.095445115, 1]):

Example 4
Write the equation of the hyperbola in standard form and determine its center, vertices, and foci.

Plot the graph, state and plot the asymtotes and Foci.

3*x^2-y^2-18*x+10*y-10 = 0

> completesquare(%,x);

Standard form: (x-3)^2/4-(y-5)^2/12 = 1.

a^2 = 4 and b^2 = 12 give c^2 = 16 .

Vertices: (3 +-2,5)

Foci: (3 +- 4,5)

Center: (3,5)

asymptotes: y = +- sqrt(12)/2*x .

> f:= x -> 5 + 1.732050808 * (x-3) ;

f := proc (x) options operator, arrow; -.196152424+...

> A:= implicitplot((x-3)^2/4-(y-5)^2/12 = 1, x = -10..12, y = -20..20):
B:= pointplot([7,5]):

> E:= plot(f(x), x = -10..10, color= blue):

> display({A,B,C,E,F});

[Maple Plot]


Example 5
Using the points (0,0), (-4,0), (0,6) and the general equation x^2+y^2+D*x+E*y+F = 0

we obtain:

F = 0

16 - 4D + F = 0

36 + 6E + F = 0.

Equation of circle is: x^2+y^2+4*x-6*y = 0 .

> completesquare(%,y);

Example 6
Write the equation of the parabola that has the given characteristics.

Plot Each parabola with Focus and directrix. Label the focus and directirx on the graph.

a) vertex (3,-2) and directrix x = -5

b) passes through (5,10), vertex is at the origin, and the axis is the x-axis.

> A:= implicitplot((y+2)^2 = 32*(x-3), x = -10..12, y = -20..20):

> C:= textplot([-7,7,'directrix'], color = blue):

> E:= textplot([11.5,-5,'focus'], color = blue):

c) We use: (y-k)^2 = 4*p*(x-h) . Vertex is (0,0) implies that h = 0 and k = 0.

Since (5,10) is on the graph we have:

100 = 4 p 5 = 20 p OR p = 5.

Thus equation is: y^2 = 20*x .

Vertex is: (0,0)

Focus is: (5,0)

Directrix is: x = -5.

> B:= plot([-5,t,t = -10..10], color = blue):

> DD:= pointplot([5,0] ):

> display({A,B,C,DD,E});

[Maple Plot]

Example 7
Write the equation of the parabola in standard form and determine its vertex, focus, directrix.

x^2-4*x+8*y+36 = 0

> completesquare(x^2 -4*x + 8*y + 36 = 0,x);

(x-2)^2+32+8*y = 0

Standard form: (x-2)^2 = -8*(y+4) .

Vertex: (2,-4)

Focus: (2,-6)

Directrix: y = -2

> A:= implicitplot(x^2-4*x+8*y+36 = 0, x = -10..12, y = -20..20):B:= plot([t,-2,t = -10..10], color = blue):

> DD:= pointplot([2,-6] ):

> display({A,B,C,DD,E});

[Maple Plot]

> completesquare(4*x^2 + 5*y^2 - 24*x - 10*y + 17 = 0, y);

5*(y-1)^2+12+4*x^2-24*x = 0

> (%)/24;

5/24*(y-1)^2+1/2+1/6*x^2-x = 0

> evalf(sqrt(6/5));

1.095445115

> B:= pointplot([3 +1.095445115 , 1] ):

> display({A,B,C});

[Maple Plot]

> completesquare(3*x^2 - y^2 - 18*x + 10*y - 10 = 0,y);

-(y-5)^2+15+3*x^2-18*x = 0

> (%)/12;

-1/12*(y-5)^2+5/4+1/4*x^2-3/2*x = 0

> evalf(sqrt(12)/2);

1.732050808

> g:= x -> 5 -1.732050808 * (x - 3) ;

g := proc (x) options operator, arrow; 10.19615242-...

> A:=implicitplot(3*x^2 - y^2 - 18*x + 10*y - 10 = 0, x = -6..10, y = -10..10, grid=[40,40]):

> C:= pointplot([-1,5] ):

> F:= plot(g(x), x = -10..10, color= blue):

> display(A,C,F);

[Maple Plot]

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