quadratic identities! + HSC 98 Q (1 Viewer)

[sf_diegoxrock]

the points (1,-2), (3,0) and (-2,10) lie on a parabola, find its eqn
i tried doing it using simulataneous eqns and didnt get the answer
any help is appreciated

and a Q from the 98 HSC paper:
what is the co ord of the vertex for x*2=8(y+3)
is it (0,-3)? and the eqn of the directrix, is it y=-2?

 

[pLuvia]

Still doing the first one

2.

x²=8(y+3)
x²=8y+24
y=x²/8 - 24/8

y=x²/8 - 3

axis of symmetry x=-b/2a

x=0
y=-3

Vertex (0,-3)

Focal length = 4a=8
a=2

Function concave up
Directrix y=-5

 

[acmilan]

y = ax² + bx + c

(1,-2)

a + b + c = -2

(3,0)

9a + 3b + c = 0

(-2,10)

4a - 2b + c = 10

Three equations to solve:

9a + 3b + c = 0___(1) -> c = -9a-3b
a + b + c = -2____(2)
4a - 2b + c = 10____(3)

Sub c into 2 and 3

a + b - 9a - 3b = -2
-8a - 2b = -2
4a + b = 1______(3)

4a - 2b - 9a - 3b = 10
-5a - 5b = 10
a + b = -10____(4)

Unless there is an error, you'll get the answer by solving 3 and 4 simultaneously to get a and b, then sub the answer into c = -9a-3b to get c

 

[PLooB]

Still doing the first one

2.

x²=8(y+3)
x²=8y+24
y=x²/8 - 24/8

y=x²/8 - 3

axis of symmetry x=-b/2a

x=0
y=-3

Vertex (0,-3)

Focal length = 4a=8
a=2

Function concave up
Directrix y=-2

2/3 marks for you

focal length is indeed 2 as 4a=8
so the directrix is the line 2 below the vertex.

Thus directrix is y=-5

At first I was thinking how can the directrix go through the parabola? Then I realised this wasn't a x²=4ay question.

 

[pLuvia]

oh yeh, sorry I'll fix it up