### 12.1 Parabolas

```SUTCLIFFE’S NOTES: CALCULUS 2 SWOKOWSKI’S CHAPTER 12
12.1 Parabolas
Definition A parabola is the set of all points in a plane equidistant from a fixed point F (the focus) and a
fixed line l (the directrix) in the plane.
To derive an equation of a parabola with vertex at the origin, focus at F  0, p  and directrix y  p . A
point P  x, y  is on the parabola if and only if
 x  02   y  p2

 x  x 2   y  p2 .
Square both sides and simplify to get
 x  02   y  p 2   x  x 2   y  p 2
x 2  y 2  2py  p2  y 2  2py  p2
x 2  4py
We can also write the equation as y 
1 2
x .
4p
If p  0 , the parabola opens upward.
If p  0 , the parabola opens downward.
The graph is symmetric with respect to the x-axis.
1 2
y . This is an equation of a
4p
parabola with vertex at the origin, focus F  p,0 , and opening right if p  0 or left if p  0 .
Interchanging the roles of x and y, we get y 2  4px , or equivalently, x 
y
1 2
x
4p
y 
x 2  4py
V (0,0), F(0, p)
1 2
x
4p
x 2  4py
V (0,0),F (0, p)
x
1 2
y
4p
y 2  4px
V (0,0),F ( p,0)
x
1 2
y
4p
y 2  4px
V (0,0),F (p,0)
Note: For any nonzero real number a, the graph of y  ax 2 or x  ay 2 is a parabola with vertex V  0,0 .
We have a 
1
4p
or p 
1
.
4a
Exercises Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and
the directrix.
#2 x  2y 2
Solution:
1
1
2p
4p
8
1
1 
F  ,0  and directrix is at x  
8
8


#4 x 2  3y
Solution:
x 2  3y  4p  3  p  
3
4
3
3

F  0,    and directrix at y 
4
4

Translation of axes formulas
If  x, y  are the coordinates of a point P in an xy-plane and if  x, y  are the coordinates of P in an x y  plane with origin at the point  h, k  of the xy-plane, then
(i) x  x  h, y  y   k
(ii) x  x  h, y   y  k
Exercises Find the vertex and focus of the parabola. Sketch its graph, showing the focus.
#8 y  8x 2  16x  10
Solution:


y  8x 2  16x  10  8 x 2  2x  1  10  8  8  x  1  2
2
V  1,2
1
1
8p
4p
32
1   65 

F  1,2     1, 
32   32 

#10 y 2  20y  100  6 x
Solution:
y 2  20y  100  6x
 y  102  6x  V 0,10
4p  6  p 
3

F  ,10 
2


3
2
or 6x   2  y  10   0  V 0,10 
#12 y 2  14y  4x  45  0
Solution:
y 2  14y  4x  45  0
y
2

 14x  49  45  49  4 x   y  7   4  4 x   y  7   4  4 x   y  7   4  x  1
2
2
Thus, V 1, 7  .
Or, get x  and set to 0 to get V .
4p  4  p  1
F  0, 7 
#14 y  40x  97  4x 2
Solution:
y  40x  97  4x 2


y  4x 2  40x  97  4 x 2  10x  25  97  100  4  x  5   3  V 5,3 
2
Or, y   8x  40  0  x  5, y  40 5   97  4 5   3  V 5,3 
2
1
1
 4  p  
4p
16
1

 47 
F  5,3    F  5, 
16


 16 
2
Exercises Find an equation of the parabola that satisfies the given conditions.
#18 focus F 0, 4 ; directrix y  4
Solution:
This is a parabola with vertex at the origin and opening downward and will have equation x 2  4py .
p  4  4p  16
Thus, an equation is x 2  16y .
#20 vertex V  2,3 ; directrix y  5
Solution:
This is a parabola that opens downward and will have equation  x  2  4p  y  3 . The distance of the
2
directrix from the vertex is 5-3) or 2 units which means that p  2 and thus an equation is
 x  22  8  y  3 .
#22 vertex V 1, 2 ; focus F 1,0
Solution:
This is a parabola that opens upward and will have equation  x  1  4p  y  2 . The distance between
2
the focus and the vertex is 0-(-2) or 2 units which means that p  2 and an equation is
 x  12  8  y  2 .
#24 vertex V  3,5 , axis parallel to the x-axis and passing through the point 5,9
Solution:
This is a parabola that opens to the right and will have equation  y  5  4p  x  3 . We can replace x
by 5 and y by 9 to get p:
1
9  52  4p 5  3  16  4p 8  p 
2
2
An equation is  y  5  2  x  3 .
2
#26 Find an equation of the parabola that has a vertical axis and passes through
A  2,5, B  2, 3, and C 1,6 .
Solution:
We can substitute the x and y values from the three points in the equation y  ax 2  bx  c to get a 3 by 3
linear system in a, b, and c.
5  4a  2b  c
3  4a  2b  c
6  abc
Solve to get a  1, b  2, and c  5 and the equation is y  x 2  2x  5.
#28 Let R be the region bounded by the parabola y 2  2x  6 and x  5 .
(a) Find the area of R.
(b) If R is revolved about the y-axis, find the volume of the resulting solid.
(c) If R is revolved about the x-axis, find the volume of the resulting solid.
Solution:
(a) Find the points of intersection of the curves:
y 2  2 5  6  4  y  2
2
y2  6 
1 3
8
8 16


A  2  5 
sq units
 dy  2 2y  y   2 4    2  
0
2
6
6
3 3

0




(b) Using washers, we get
2
2
2
2
 y2  6  
1 4
1 4

2
2
V  2  52  
  dy  2 0 25   y  24y  36  dy  2 0 16  y  6y  dy
0
2
4
4







 
2
1 5
224


 2 16y 
y  2y 3  
cubic units
20
5

0
(c) We can revolve the top half and using cylindrical disks, we get
2
V     2x  6  dx    x 2  6x 3   25  30  9  18  4 cubic units
5
5
3
#34. Prove that the point on a parabola that is closest to the focus is the vertex.
Proof:
Let x 2  4py be a parabola with vertex  0,0 and focus  0, p  .

Let  x,

x2 
 be any point on the parabola.
4p 

Want: minimize the distance between  x,

x2 
 and  0, p  .
4p 
We can also minimize the square of the distance to get:
2

 x2
 p
D   x  0  

 4p
2
 2x
x
0
 p 
D  2x  2 
 4p
 4p
2
2x 
2x
2x 2 2x
0
 2p 

4p
4p 4p
1+
x2 1
 0
8 p2 2
1
x2

2
2
8p
x 2  4 p2
no solution
or
x 0
```