Download CONIC SECTIONS Problem 1 For the equations below, identify each

CONIC SECTIONS
Problem 1
For the equations below, identify each conic section. If it’s a parabola, specify its vertex, focus and
directrix. If it’s an ellipse, specify its center, vertices and foci. If it’s a hyperbola, specify its center,
vertices, foci and asymptotes.
x 2  3 y 2  12 y  9
a)
 x 2  3 y 2  12 y  9
 x 2  3  y 2  4 y   9
 x 2  3  y  2   12  9
2
 x2  3  y  2  3
2

x2
 3
2
 y  2

12
2
1
So the conic is an ELLIPSE, where a  3 , b  1 , and c  3  1  2 . We therefore conclude
the following:
b)

Its center is the point C   0, 2 .

Its major axis is the horizontal line y  2 and its minor axis is the y -axis.

Its vertices are located at the points V1 

Its foci are located at the points F1 
8 y  4   x  1
2
1
2

 8  y     x  1
2

1
2

  x  1  4  2   y  
2






3, 2 and V2   3, 2 .



2, 2 and F2   2, 2 .
So the conic is a PARABOLA that OPENS UP, where a  2 . We therefore conclude the
following:



1

Its vertex is located at the point V  1,   .
2

 3
Its focus is located at the point F  1,  .
 2
5
Its directrix is the horizontal line y   .
2
9  x  1  36  4 y 2
2
c)
 9  x  1  4 y 2  36
2

 x  1
2
22
y2
 2 1
3
So the conic is a HYPERBOLA, where a  2 , b  3, and c  9  4  13 . We therefore
conclude the following:



Its center is the point C 1,0.

Its transverse axis is the x -axis and its conjugate axis is the vertical line x  1 .
Its vertices are located at the points V1  3,0 and V2  1,0.

Its foci are located
at the points F1  1  13, 0 and F2  1  13, 0 .



3
3
3
3
Its asymptotes are the lines y  x  and y   x  .
 2

2
2
2

d)

2 x2  3 y 2  4 x  6 y  13
 2  x 2  2 x   3  y 2  2 y   13
 2  x  1  3  y  1  13  2  3  18
2

 x  1
32
2
2
 y  1

 
6
2
2
1



So the conic is an ELLIPSE, where a  3 , b  6 , and c  9  6  3 . We therefore conclude
the following:



Its center is the point
 1,1.
 C

Its major axis is the horizontal line y  1 and its minor axis is the vertical line x  1 .
Its vertices are located at the points V1  4,1 and V2  2,1.

Its foci are located
at the points F1  1  3,1 and F2  1 3,1 .





x2  2 y 2  3  2 x  8 y
e)






 x2  2x  2 y 2  8 y  3
  x  1  2  y 2  4 y   3  1  4
2
  x  1  2  y  2   4  8  4
2
2
 2  y  2    x  1  4
2

 y  2
 2
2
2
2
 x  1

22
2
1
So the conic is a HYPERBOLA, where a  2 , b  2, and c  4  2  6 . We therefore
conclude the following:



Its center is the point C  1,2.



Its transverse axis is the vertical line x  1 and its conjugate axis is the horizontal line
y  2 .


Its vertices are
 located at the points V1  1,2  2 and V2  1,2  2 .

Its foci are located at the points F1  1,2  6 and F2  1,2  6 .

Its asymptotes are the linesy 









 2 4
 2 4
2
2
x  
y


x

and


 .

2
2
 2 
 2 

Problem 2
[Exercise # 72 on page 645: Reflecting Telescope]
Since the mirror of the telescope is shaped like a paraboloid of revolution (i.e. a parabola
rotated about its axis of symmetry), the collected light will be concentrated at the focus point
of the parabola (technically, it’s the focus point of any cross-section of the paraboloid of
revolution passing through its vertex).
Therefore, we seek the y-coordinate of the focus point of a parabola that opens up, and has its
vertex point at the origin of a Cartesian plane. Moreover, given the dimensions of the mirror,
we know that the points  2,3 and  2,3 belong to the parabola [do you see why?]
All of this implies that the equation of the parabola is given by x 2  4ay , where a  0 , and that
the focus is located at the point  0, a  , or a inches above the vertex point. Plugging in the
coordinates of the point
which in turn yields a 
 2,3
into the equation of the parabola yields  22   4a(3)  12a ,
4 1
  0.33 .
12 3
We conclude that the light will be concentrated approximately 0.33 inches above the
vertex point of the paraboloid of revolution.
Problem 3
[Exercise # 82 p. 669: Equilateral Hyperbola]
We know that the eccentricity of an ellipse, denoted by e , is defined as the positive number
c
.
a
We also know that if the hyperbola is equilateral, then a  b .
Since b 2  c 2  a 2 for any hyperbola, we conclude that c  b2  a2  2a2  2a for an
equilateral hyperbola. Therefore its eccentricity is given by e 
c
2a


a
a
2.
Problem 4
[Exercise # 72 on page 655: Whispering Gallery]
Here the whispering gallery is shaped like the upper portion of an ellipse whose foci are located
100 feet from each other and whose vertices are located 6 feet from the foci. This implies that
if we place the center of the ellipse at the origin, then the foci are located at the points
 50,0 and the vertices are located at the points  56,0 .
We then have c  50 , a  56 , and b  a2  c2  562  502  25.2 .
Therefore, the gallery is 112 feet long and has a maximal height (at the center of the room)
of approximately 25.2 feet.
Problem 5
[Exercise # 78 on page 655: Volume of a Football]
First, note that a longitudinal cross-section of the prolate spheroid (i.e. the football) passing
through its center is an ellipse in the Cartesian plane whose center is at the origin, whose
major axis is the x-axis, and whose vertices and foci lie on the x-axis.
Since the football is 11.125 inches in length, the vertices of the ellipse are located at the points
 5.5625,0 and 5.5625,0 , and so we have a  5.5625 .
Since the football has a center circumference of 28.25 inches, we have 2 b  28.25 [do you see
28.25
 4.4961 .
why?] This implies that b 
2
We conclude that the volume of the football, in cubic inches, is given by
4
2
V    5.5625  4.4961  471.01
3
Problem 6
a)
Identify the conic section represented by the polar equation r  3  2r cos .
r  3  2r cos 
 r  2r cos   3
 r1  2cos    3
3 
2 
3
2 
r

1  2cos  1  2cos 
Therefore, e  2  1 and the conic is a hyperbola.

c)

Find the rectangular equation of the conic. Justify your work analytically.
If r  32rcos , then we can convert the polar equation of the hyperbola to
rectangular coordinates as follows:

r 2   3  2r cos  
2
 x2  y 2  3  2x 
2
 x 2  y 2  9  12 x  4 x 2
 y 2  3x 2  12 x  9
 y 2  3 x2  4x   9
 y 2  3  x  2   12  9
2
 y 2  3  x  2   3
2
 3 x  2  y2  3
2

 x  2
12
2

y2
 3
2
1
PARAMETRIC EQUATIONS
Problem 1
Consider the plane curve C defined by the following parametric equations:
x  4sin t  2sin(2t )
y  4 cos t  2 cos(2t )
b)
,
(t R)
What type of graph do you recognize here?
A cardioid.
c)
Find the polar equation of C . Justify your work analytically.
If x  4 sin( t)  2sin( 2t) and y  4 cos(t)  2cos(2t) , then we have
r2  x 2  y 2
 16sin 2
(t)  cos2 (t) 4 sin 2 (2t)  cos 2 (2t)16sin( t)sin( 2t)  cos(t)cos(2t)

16(1)  4(1) 16sin( t)sin( 2t)  cos(t)cos(2t)


 20 16 2sin 2 (t)cos(t)  cos(t)1 2sin 2 (t)
 20 16cos(t)



The polar equation of C is therefore given by


r 2  20 16cos(t)

Problem 2
[Exercise # 54 p. 696: Projectile Motion]
Here we have v0  125 ft / sec ,   400 , and h  3 ft .
a)
The parametric equations that model the position of the ball are given by
x  x(t )   v0 cos   t  125cos  400  t
1
y  y (t )   gt 2   v0 sin   t  h  16t 2  125sin  400   t  3
2
b)
To find out how long the ball is in the air, find the positive root to the quadratic
equation y  0 :
y0
 16t 2  125sin  400   t  3  0
 125sin  400    125sin  400    4  16  3
t
2  16 
2
 t  5.06sec
So the ball traveled for a total of 5.06 seconds.
c)
To determine the horizontal distance that the ball traveled, compute x(5.06) :
x(5.06)  125cos  400   5.06   484.52 ft
So the ball traveled an approximate total horizontal distance of 484.52 feet.
d)
To determine the maximal height attained by the ball, determine the y-coordinate of the
 b 
vertex point of the parabola y  y (t ) , or y    :
 2a 
 b 
y  
 2a 
 125sin 400 
y
  y  2.51  103.87 ft .
2(16) 

So the ball reached a maximal height of approximately 103.87 feet.