1. LECTURE UNIT 004
Parabola
The locus of a moving point that its distance from a fix point ( ) is equal to its distance from a fixed
line ( ).
y
x = -a
Directrix
d1
2a
Latus Rectum 4a
d2
Where:
x
A v(h, k) F F = Focus
2a
v = Vertex, midpoint of the segment
d 1 = d2
e = eccentricity = 1
a a
2a
If the vertex (h, k) is at the origin, the equation of the parabola is:
y2 = 4ax ( )
If the vertex is a horizontal axis parabola ( ) the equation of the parabola is:
(y - k)2 = 4a (x - h)
If the vertex (h, k) is at the origin, the equation of the parabola is:
x2 = 4ay ( )
If the vertex is a vertical axis parabola ( ) the equation of the parabola is:
(x - h)2 = 4a (y - k)
Cases:
y y
y y
x=-a x=a
F
y=a
v
x x v
x x
v v
F F
y=-a
F
(a) y2 = 4ax (b) y2 = -4ax (c) x2 = 4ay (d) x2 = -4ay
Sketch the graph:
1. (y - 2)2 = 4 (x + 3)
2. x2 = -8 (y + 1)
3. (y + 2)2 = -6 (x - 1)
“The only active force that arises out of possession is fear of losing the
object of possession!”

2. The equation x2 + Dx + Ey + F = 0 is a parabola with vertical axis (horizontal directrix) and the equation
y2 + Dx + Ey + F = 0 is a parabola with horizontal axis (vertical directrix). To sketch the graph, reduce the
equation to standard form.
4. y2 - 2x = 0
5. x2 + 12y - 2x - 11 = 0
6. 2y2 - 5x + 4y - 7= 0
7. y2 + 4x - 6y + 1 = 0
8. x2 - 2y - x= 0
9. y2 - 7x + 3y - 8 = 0
10. x2 - 8y + 4x - 4= 0
11. (x + 2)2 = - (y - 1)
12. y2 - 4x + 2y - 7 = 0
13. x2 + 2y - x = 0
Determine the points of intersection and sketch the graph on the same axis.
14. y = x2 + 3x and y = x + 3
15. y = x2 - 3 and y = x2 + 5
16. y = x2 + 5x - 5 and y = x
17. y2 = 2x and x2 = 2y
18. y = x2 and y = 2x + 3
Find the equation of the parabola with given conditions.
19. With directrix y = 2 and a focus at (-2, 4).
20. With vertex at the intersection of 3x - y = 7 and x - 2y = 4 and directrix is x = 4.
21. With vertex at the center of the circle x2 + y2 - 3x - 2y = 0 and passing through (-2, -3) with vertical axis.
22. With vertex at the origin and passing through (1, -3) with horizontal, find the length of the latus rectum.
“A house is the character of people who live in it”

3. PARABOLA
Directrix (x=2.5)
Example 3:
Sketch the graph of (y + 2)2 = -6 (x - 1)
Solution:
(y + 2)2 = -6 (x - 1)
4|a| = 6 2a
2|a| = 3
a a
F
|a| = 1.5 v(1,-2)
v(1, -2) parabola opens to the left
2a
Example 6:
2y2 - 5x + 4y - 7= 0
Solution:
Reducing to Standard Form:
[2y2 - 5x + 4y - 7= 0] 1 Multiplying both sides of the equation by 1/2
2
2
y - 5 x +2y - 7 = 0
2 2
Directrix (x=2.42)
y2 + 2y = 5 x + 7
2 2
Completing squares;
y2 + 2y + 1 = 5 x + 7 + 1
2 2
(y + 1)2 = 5 x + 9 v(-1.8,-1) F
2 2
Factoring x;
9
(y + 1)2 = 5 (x + )
2 5
4|a| = 2.5
2|a| = 1.25
|a| = 0.625
v (-1.8, -1) parabola opens to the right
Example 14:
y = x2 + 3x and y = x + 3
Solution:
By substitution;
X2 + 3x = x + 3
X2 + 2x + 3 = 0
(x + 3)(x - 1) = 0
Points of intersection;
P1 (-3, 0) and P2 (1, 4)
Graph of the parabola;
x2 + 3x + 9 = y + 9
4 4
2
(x + 3 ) = y + 9
2 4
v (-1.5, -2.25) Parabola opens upward
“For every joy there is a price to be paid”

4. Graph of the line;
x y
+ =1
-3 3
P2(1, 4)
(0, 3)
P1(-3, 0)
v(-1.5,-2.25)
Example 17:
y2 = 2x and x2 = 2y
Solution: x2 = 2y
By substitution; (2, 2)
2
x2
( 2 ) = 2x
Points of intersection;
P1 (0, 0) and P2 (2, 2)
Graph of the parabola; v(0, 0)
y2 = 2x
v(0, 0) parabola opens to the right
2
x = 2y
v(0, 0) parabola opens upward
y2 = 2x
Example 21:
With vertex at the center of the circle x2 + y2 - 3x - 2y = 0 and passing through (-2, -3) with vertical axis.
Solution:
C(- D ,- E )
2 2
C( 3 ,1)
2
Note that the vertex is at the center of the circle
(x - h)2 = 4a (y - k)
2
(x - 3 ) = 4a (y - 1)
2
Solving for a, using P (-2, -3)
2
(-2 - 3 ) = 4a (-3 - 1)
2
a = - 49
64
Therefore;
2
(x - 3 ) = - 16 (y - 1)
2
49
parabola opens downward
“The best and shortest road towards knowledge of truth is nature”