1. Parabola – Locus II
By Mr Porter

2. Assumed Knowledge: Simple Examples
Students need to know, understand and apply the skills to complete the square of a quadratic.
Simple Examples
a) x2 + 8x + 12 = 0
Rearrange to x2 + bx = c b)
Divide by 2 and rearrange to x2 + bx = c
x2 + 8x = -12
For ax2 + 2bx + b2 = 0
2b = (+8)
For ax2 + 2bx + b2 = 0
2b = (+4)
To complete the square we need to add b2 = +16 to both sides.
To complete the square we need to add b2 = +4 to both sides.
x2 + 8x + 16 =
x2 + 8x + 16 = 4
Factorise LHS.
(x + 4)2 = 4
Take √ of both sides.
Take √ of both sides.
x + 4 = ±2
Solve for x.
Solve for x.
x = -4±2
i.e. x = or x =
x = -6 or x = -2
The solution is x = -6 and x = -2.

3. Example 1:
Rewrite y = x2 – 4x + 6 in the form (x – h)2 = 4a(y – k) and hence state the vertex, focus and directrix of the parabola.
Solution:
So, Vertex (h, k) = (2, 2), focal length a = ¼.
Rearrange with all x-terms on the right and everything else on the left side of the ‘=‘.
Focus (h, k +a) = (2, 2 + ¼)
x2 – 4x + 6 = y
Parabola is concave-up!
x2 – 4x = y – 6
Now, “complete the square” for ‘x’.
Directrix y = k – a
x2 – 4x + 4 = y – 6 + 4
Factorise RHS.
y = 2 – ¼
(x – 2)2= y – 2
Write in the form
(x – h)2 = 4a(y – k)
(x – 2)2= 1(y – 2)
(x – h)2 = 4a(y – k), vertex at (h, k)

4. Example 2:
Rewrite y = 2x2 + 8x – 10 in the form (x – h)2 = 4a(y – k) and hence state the vertex, focus and directrix of the parabola.
Solution:
So, Vertex (h, k) = (-2, -18),
focal length a = 1/8, concave up (+).
Rearrange with all x-terms on the left and everything else on the right side of the ‘=‘.
Focus (h, k +a) = (-2, /8)
2x2 + 8x – 10 = y
Parabola is concave-up!
2x2 + 8x = y + 10
Now, “complete the square” for ‘x’.
Directrix y = k – a
x2 + 4x = 1/2 y +5
y = -18 – 1/8
x2 + 4x + 4 = 1/2 y
Factorise LHS.
(x + 2)2= 1/2 y + 9
Write in the form
(x – h)2 = 4a(y – k)
(x + 2)2= 1/2 (y + 18)
(x – -2)2= 1/2 (y – -18)
(x – h)2 = 4a(y – k), Vertex at (h, k)

5. Example 3:
For the parabola, y = –x2 + 6x – 10 , state the vertex, focus and write the equation of the directrix.
Solution:
So, Vertex (h, k) = (3, -1),
focal length a = 1/4,, concave down (-).
Rearrange with all x-terms on the left and everything else on the right side of the ‘=‘.
Focus (h, k +a) = (3, -1 – 1/4)
–x2 + 6x – 10 = y
Parabola is concave-down!
x2 – 6x = –y – 10
Rearrange and “complete the square” for ‘x’.
Directrix y = k + a
x2 – 6x + 9 = –y –
y = /4
x2 – 6x + 9 = –y – 1
Factorise LHS.
(x – 3)2 = –y – 1
Write in the form
(x – h)2 = –4a(y – k)
(x – 3)2 = –(y + 1)
(x – 3)2 = –(y – -1)
(x – h)2 = -4a(y – k), Vertex at (h, k)

6. Example 4:
For the parabola, y = – 1/4 x2 + 2x – 2 , state the vertex, focus and write the equation of the directrix and the length of the latus rectum.
Solution:
So, Vertex (h, k) = (4, 1),
focal length a = 1, concave down (-).
Rearrange with all x-terms on the left and everything else on the right side of the ‘=‘.
Parabola is concave-down!Multiply everything by ‘4’.
Focus (h, k +a) = (4, 1 – 1)
–¼ x2 + 2x – 2 = y
= (4, 0)
x2 – 8x = –4y – 8
Rearrange and “complete the square” for ‘x’.
Directrix y = k + a
x2 – 8x = –4y –
y = 1 + 1
x2 – 8x = –4y + 8
Factorise LHS.
y = 2
(x – 4)2 = –4y + 8
Write in the form
(x – h)2 = –4a(y – k)
Length of latus rectum = |4a|
(x – 4)2 = –4(y – 1)
= 4
(x – h)2 = -4a(y – k), Vertex at (h, k)

7. Example 5:
For the parabola, x = y2 + 2y – 7 , state the vertex, focus and write the equation of the directrix and the length of the latus rectum.
Solution:
So, Vertex (h, k) = (-1, -8),
focal length a = ¼, concave right (+) .
Rearrange with all y-terms on the right and everything else on the left side of the ‘=‘.
Focus (h +a, k) = (-1 + ¼ , -8)
y2 + 2y – 7 = x
Parabola is concave-right!
= (-3/4 , -8)
y2 + 2y = x + 7
Rearrange and “complete the square” for ‘y’.
Directrix x = h – a
y2 + 2y + 1 = x
x = -1 – ¼
y2 + 2y + 1 = x + 8
Factorise LHS.
x = - 5/4
(y +1)2 = x + 8
Write in the form
(y – k)2 = 4a(x – h)
4x + 5 = 0
(y + 1)2 = 1(x + 8)
(y – -1)2 = 1(x – -8)
Length of latus rectum = |4a|
= 1
(y – k)2 = 4a(x – h), vertex at (h, k)

8. Example 6:
For the parabola, x = – 1/3 y2 – 2y + 9 , state the vertex, focus and write the equation of the directrix and the length of the latus rectum.
Solution:
So, Vertex (h, k) = (-3, -12),
Focal length ‘a’ = 3/4 , concave left (-)
Rearrange with all y-terms on the right and everything else on the left side of the ‘=‘.
Parabola is concave-left!
Multiply everthing by ‘-3’.
Focus (h + a, k) = (-3 + – 3/4 , -12)
–1/3 y2 – 2y + 9 = x
= (-33/4 , -12)
y2 + 6y = -3x + 27
Rearrange and “complete the square” for ‘y’.
Directrix x = h – a
y2 + 6y + 9 = -3x
x = -3 – - 3/4
y2 + 6y + 9 = -3x + 36
Factorise LHS.
x = -15/4
(y + 3)2 = -3(x + 12)
Write in the form
(y – k)2 = -4a(x – h)
4x + 15 = 0
(y + 3)2 = -3(x + 12)
(y + -3)2 = -4( 3/4 )(x – -12)
Length of latus rectum = |4a|
= 3
(y – k)2 = 4a(x – h), vertex at (h, k)