Conics Extensions
Name_________________________________
Eccentricity
Period_________Date___________________
Eccentricity is defined to be a constant expressed as the ratio of the distance from a point on a
conic to a focus and the distance from the point to the directrix. That means that parabolas
aren’t the only conics with a directrix - hyperbolas and ellipses have 2 of them, one for each
focus, as partially shown in the figures below:
The equations for the directrices come from the
definition of eccentricity:
Call the distance from the center to the directrix d.
c a c

a d a
At the close endpoint of the major axis,
Cross multiply to get cd – ca = a2 – ca  cd = a2
So d = a2/c, making the equation of this directrix x = a2/c.
We can use this information to solve problems like this:
Example: A conic has focus at (3, 4), directrix at
x = -2 and eccentricity of 2/3. Write the equation of this
conic in standard form.
Solution: Let (x, y) be any point on the conic. Using the definition
 x  3  y  4 
2
PF
of eccentricity:

Pd
2
x 2
Cross multiply to get: 3

 x  3  y  4
2
2
3
2

 2  x  2


9 x 2  6x  9  y 2  8y  16  4 x 2  4x  4

Square both sides: 9x  54x  81  9y  72y  144  4x  16x  16
2
2
2
5x 2  9y 2  70x  72y  209  0
5(x 2  14x  49)  9(y 2  8y  16)  209  245  144  180
5(x 2  14x  49) 9(y 2  8y  16)

1
180
180
Complete squares: (x 2  14x  49) (y 2  8y  16)

1
36
20
(x  7)2 (y  4)2

1
36
20
So this ellipse has center at (7, 4). The endpoints of the major axis are (1, 4) and (13, 7) and
the foci are at (3, 4) and (11, 7). And to check, e = c/a = 4/6 = 2/3. 
Try these:
I. For each of the following, A) name the type of conic section, B) write its equation in standard
form, and C) sketch the graph, showing the focus and directrix.
1. This conic has a focus is at (0, 0), corresponding directrix at y = 3 and eccentricity of 2.
2. This conic has a focus is at (0, 0), corresponding directrix at x = 1 and eccentricity of 0.6 .
3. This conic has a focus is at (0, 0), corresponding directrix at x = -4 and eccentricity of 1.
4. This conic has foci at (-3, 8) and (-3, -4) and eccentricity of 1.2 .
5. This conic has foci at (6, 10) and (6, -14) and eccentricity of 12/13.
6. This conic has a center at (4, 7), one vertex at (21, 7) and eccentricity of 15/17.
7. This conic has a center at (1, 3), directrix at y = 5 and corresponding focus at (1, 10).
8. This conic has a center at (1, 2), directrix at x = 7 and corresponding focus at (3, 2).
9. This conic has vertices at (3, 9) and (13, 9) and eccentricity 1/5.
10. This conic has vertices at (3, 9) and (13, 9) and eccentricity 5.
11. This conic only has only one vertex at (4, 8) and a focus at (0, 8).
12. This conic has an eccentricity of 1/3 and the endpoints of its minor axis are (4, 8) and (0, 8).
13. This conic has only one vertex at (3, 2) and a directrix at x = 0.
14. This conic has an eccentricity of 3/2, a focus at (6, 3) and asymptotes of y – 3 = ±2x .
II. Consider the ellipse:
x2 y2

 1 . There exists a hyperbola whose vertices are the foci of
25 9
this ellipse and whose foci are the vertices of this ellipse. Write the equation of this hyperbola
and write the eccentricities of both figures.
Try these:
I. For each of the following, A) name the type of conic section, B) write its equation in standard
form, and C) sketch the graph, showing the focus and directrix.
1. This conic has a focus is at (0, 0), corresponding directrix at y = 3 and eccentricity of 2.
A) It is a hyperbola because the eccentricity is greater than 1.
x2  y2
3y
B)

 2  x 2  y 2  2 3  y   x 2  y 2  4 9  6y  y 2
C)

x 2  y 2  36  24y  4y 2  x 2  3y 2  24y  36 

directrix

x 2  3 y 2  8y  16  36  48  12
y  4   x 2  1
x 2 3 y  4 

1
12
12
4
12
2
y=3
2
Focus (0, 0)
2. This conic has a focus is at (0, 0), corresponding directrix at x = 1 and eccentricity of 0.6 .
A) It is an ellipse because the eccentricity is less than 1.
3
 5 x 2  y 2  3 1  x   25 x 2  y 2  9 1  2x  x 2
1x
5
2
2
25x  25y  9  18x  9x 2  16x 2  25y 2  18x  9 

2
B) . 16  x 






9
81 
81 225
2
x

  25 y  9 
8
256 
16
16
 
2
2


9
9
2
16  x  
x 

25
y
   1   16   y 2  1
16 


9
225
225
225
16
16
256
Focus (0, 0)
16
3. This conic has a focus is at (0, 0), corresponding directrix at x = -4 and eccentricity of 1.
A) This figure is a parabola because the eccentricity is 1.
B) Since the vertex is halfway between the focus and the directrix,
it must be at (-2, 0). Since the vertex is to the left of the focus,
the parabola opens right and since the distance between the vertex
and the focus is 2, p = 2. Therefore the equation is: y2 = 8(x + 2)
C)
Directrix x = 1
(0, 0)
x2  y2
C)
4. This conic has foci at (-3, 8) and (-3, -4) and eccentricity of 1.2 .
C)
A) Since e = 1.2, this figure is a hyperbola.
B) The center is the midpoint of the foci so it is (-3, 2).
Furthermore, since the foci are above and below the center,
the hyperbola opens up/down, so the y term is positive.
Furthermore, this means c = 6 because that’s the distance
from the center to a focus. Since e = c/a , 1.2 = 6/a  a = 6/1.2 = 5.
Since a2 + b2 = c2, 25 + b2 = 36  b2 = 11. This makes the equation:
y  2    x  3
2
25
2
1
11
and d =
a 2 25
so the asymptotes are

c
6
Y = 2 + 25/6 = 37/6 and y = 2 – 25/6 = -13/6
5. This conic has foci at (6, 10) and (6, -14) and eccentricity of 12/13.
A) Since e = 12/13, the figure is an ellipse.
C)
B) The center is (6, -2) and it opens up/down, so the major
axis is parallel to the y axis. c = 12, so a = 13. Since a2 = b2 + c2,
b2 = 169 = 144 = 25. That makes the equation:
 x  6
2
25
Y = 2 
y  2 

2
169
a 2 169

so the asymptotes area
c
12
 1 and d =
169
or y =145/12 or y = -193/12
12
6. This conic has a center at (4, 7), one vertex at (21, 7) and eccentricity of 15/17.
A) This is an ellipse.
C)
B) It is longer in the x direction and a = 17 so c = 15.
b2 = 172 – 152 = 64
x - 4 
2
So the equation is
289
y - 7 

2
64
1
Foci are at (19, 7) and (-11, 7) and d = 289/15, so
Equations of asymptotes are x = 349/15 and x = -229/15
7. This conic has a center at (1, 3), directrix at y = 5 and corresponding focus at (1, 10).
A) Directrix is closer to center than focus, so it’s a hyperbola.
B) c = 7 and d = a2/c = 2 a2 = 14, a2 + b2 = c2 b2 = 49 – 14 = 35
y  3    x  1 
2
So equation is
14
2
35
1
C)
8. This conic has a center at (1, 2), directrix at x = 7 and corresponding focus at (3, 2).
A) Focus is closer to center than directrix so it’s an ellipse.
 x  1
2
B)
12
y  2 

2
1
8
9. This conic has vertices at (3, 9) and (13, 9) and eccentricity 1/5.
A) Ellipse
 x  8
2
B)
25
y  9 

2
24
1
10. This conic has vertices at (3, 9) and (13, 9) and eccentricity 5.
A) Hyperbola
 x  8   y  9 
2
B)
25
2
600
1
11. This conic only has only one vertex at (4, 8) and a focus at (0, 8).
A) Parabola
B) -16(x – 4) = (y – 8)2
12. This conic has an eccentricity of 1/3 and the endpoints of its minor axis are (4, 8) and (0, 8).
A) Ellipse
B) Center is (2, 8) , b = 2 c/a = 1/3, so a = 3c. a2 = b2 + c2  (3c)2 = 4 + c2  8c2 = 4  c2 = ½
 x  2   y  8 
So equation is:
2
a2 = 9(1/2) = 9/2.
4
2
9
2
1
13. This conic has only one vertex at (3, 2) and a directrix at x = 0.
A) Paarabola
B) 12(x – 3) = (y – 2)2
14. This conic has an eccentricity of 3/2, a focus at (6, 3) and asymptotes of y – 3 = ±2x .
A) hyperbola
B) Center (0, 3) and c = 6 and since c/a = 3/2 = 6/a  a = 4 and a2 = 16. 16 + b2 = 36  b2 = 20.
 x   y  3
2
So the equation is:
16
2
20
1
II. Consider the ellipse:
x2 y2

 1 . There exists a hyperbola whose vertices are the foci of
25 9
this ellipse and whose foci are the vertices of this ellipse. Write the equation of this hyperbola
and write the eccentricities of both figures.
In this ellipse, a2 = 25 and c2 = 16. If a hyperbola has c2 = 25 and a2 = 16, then b2 would have to
x2 y2
equal 9. Therefore the equation should be:

1
16 9