1. Parabola – Locus
By Mr Porter

2. Parabola x2 = 4ay Definition:
The locus of a point which moves sot that its distance from a fixed point (focus) is equal to its distance from a fixed straight line (directrix) is a parabola.
Features of a parabola x2 = 4ay, (a > 0). ‘a’ = focal length.
You need to know these features.

3. Example 1: x2 = 4ay, vertex (0,0) { Vertical Parabolas}
For each of the following parabolas, write down the focal length, the coordinates of the vertex, the equation of the directrix and the coordinates of the focus.
a) x2 = 8y
b) x2 = -12y
The negative sign indicates the parabola is concave down.
For x2 = 4ay, we have
For x2 = 4ay, we have
4a = 8
4a = -12
a = 2, the focal length.
a = -3, the focal length of 3 units.
Using the the focal length, a = 2
coordinates of the vertex (0, 0)
the equation of the directrix y = -a
y = -2
coordinates of the focus (0, a) = (0, 2)
Using the the focal length, a = -3
coordinates of the vertex (0, 0)
the equation of the directrix y = -a
y = 3
coordinates of the focus (0, a) = (0, -3)

4. Example 2: y2 = 4ax, vertex (0,0) {Horizontal Parabolas}
For each of the following parabolas, write down the focal length, the coordinates of the vertex, the equation of the directrix and the coordinates of the focus.
a) y2 = 18x
The posiative sign indicates the parabola is concave right
b) y2 = -10x
The negative sign indicates the parabola is concave left
For y2 = 4ax, we have
For y2 = 4ax, we have
4a = 18
4a = -10
a = 4.5, the focal length.
a = -2.5, the focal length of 2.5 units.
Using the the focal length, a = 4.5
coordinates of the vertex (0, 0)
the equation of the directrix x = -a
x = -4.5
coordinates of the focus (a, 0) = (4.5,02)
Using the the focal length, a = -2.5
coordinates of the vertex (0, 0)
the equation of the directrix x = -a
x = 2.5
coordinates of the focus (a, 0) = (-2.5, 0)

5. General Parabola with vertex at (h, k).
Equations of the general parabola with vertex at (h, k):
Vertical:
(x – h)2 = 4a(y – k), concave up.
(x – h)2 = -4a(y – k), concave down.
Horizontal:
(y – h)2 = 4a(x – k), concave right.
(y – h)2 = -4a(x – k), concave left.
‘a’ = focal length

6. Example 3
Write down the focal length, the coordinates of the vertex, the equation of the axis of symmetry, the coordinates of the focus, and the equation of the directrix of the parabola (x + 1)2 = 8(y – 4). Hence, sketch the parabola.
Writing (x + 1)2 = 8(y – 4) in the form (x – h)2 = 4a(y – k), vertex at (h, k)
Focal length: 4a = 8
sketch
a = 2
Length is 2 units.
Y-axis
X-axis
x = -1
Vertex (h, k) = (-1, 4)
Equation of the axis of symmetry: x = h
F(-1, 6)
x = -1
V(-1, 4)
focus (h, k + a) = (-1, 4 + 2)
Directrix y =2
= (-1, 6)
Equation of directrix: y =k - a
y =4 - 2
y =2

7. Example 4
Write down the focal length, the coordinates of the vertex, the equation of the axis of symmetry, the coordinates of the focus, and the equation of the directrix of the parabola (x – 2)2 = -2(y – 4). Hence, sketch the parabola.
Writing (x – 2)2 = -2(y – 4) in the form (x – h)2 = 4a(y – k), vertex at (h, k)
Note: the parabola is concave down!
Focal length: 4a = -2
sketch
x = 2
a = -0.5
Length is 0.5 units.
Y-axis
X-axis
Vertex (h, k) = (2, 4)
Equation of the axis of symmetry: x = h
Directrix y =4.5
V(2, 4)
x = 2
focus (h, k + a) = (2, )
F(2, 3.5)
= (2, 3.5)
Equation of directrix: y =k - a
y =
y =4.5

8. Example 5
Write down the focal length, the coordinates of the vertex, the equation of the axis of symmetry, the coordinates of the focus, and the equation of the directrix of the parabola (y – 3)2 = 6( x + 2). Hence, sketch the parabola.
Writing (y – 3)2 = 6(x + 2) in the form (y – k)2 = 4a(x – h), vertex at (h, k)
Note: the parabola is concave to the right!
Focal length: 4a = 6
sketch
a = 1.5
Length is 1.5 units.
Y-axis
X-axis
Vertex (h, k) = (-2, 3)
Directrix : x = -3.5
Equation of the axis of symmetry: y = k
y = 3
y = 3
V(-2,3)
F(-0.5,3)
focus (h + a, k) = ( , 3)
= (-0.5, 3)
Equation of directrix: x =h - a
x =
x =-3.5

9. Example 6
Write down the focal length, the coordinates of the vertex, the equation of the axis of symmetry, the coordinates of the focus, and the equation of the directrix of the parabola (y + 1)2 = -8( x – 3). Hence, sketch the parabola.
Writing (y + 1)2 = -8(x – 3) in the form (y – k)2 = 4a(x – h), vertex at (h, k)
Note: the parabola is concave to the left!
Focal length: 4a = -8
sketch
a = -2
Length is 2 units.
Y-axis
X-axis
Directrix : x = 5
Vertex (h, k) = (3, -1)
Equation of the axis of symmetry: y = k
y = -1
V(3,-1)
focus (h + a, k) = (3 + -2, -1)
y = -1
= (1, -1)
F(1,-1)
Equation of directrix: x =h - a
x =3 - -2
x =5