# Homework Help: Equation of a parabola

1. Nov 11, 2009

### Jshua Monkoe

1. The problem statement, all variables and given/known data

i am given the equation of a parabola to be y=2x^2-2x+3 and asked to sketch the parabola

2. Relevant equations
y=1/2(l-p)(x-k)^2+(l+p)/2
(l+p)/2=h
vertex is at (k,h)
equation of the directrix is y=p
distance[(k,h) to y=p]=distance[(x,y) to (k,h)]

3. The attempt at a solution

(1) completing the squares for y=2x^2-2x+3 i get
y=2(x-1/2)^2+2
=>k=1/2 &
h=2 &
a=2

.'. vertex is at (1,2)

(2) (l+p)/2=h=2 ........eqn1
1/(2(l-p))=a=2....eqn2

making l the subject in eqn.1 i get
l=4-p
substituting in eqn2 i get
1/(2(4-p-p))=2
=>1/(8-4p)=2
=>1=16-8p
.'.p=-15/8~-1.08
=>eqn of directrix is y=-1

(3)given the vertex is at (1,2)
F is at (1,3) taking (1,-1) from the directrix
from (2,4) to F is sq. root of 2 units
from (2,4) to (1,-1) is sq. root of 26 units
AND THIS ISNT A PARABOLA
PLEASE, WHERE HAVE I WENT WRONG BUDDIES?!?

2. Nov 11, 2009

### tiny-tim

Welcome to PF!

Hi Jshua Monkoe! Welcome to PF!

(try using the X2 tag just above the Reply box )
Nooooo

3. Nov 11, 2009

### Staff: Mentor

You have y = 2(x2 - x + ?) + 3. Remember that whatever you add in to complete the square is multiplied by 2, so you'll need to take that into account and subtract the same amount from three to keep your expressions equal.

4. Nov 11, 2009

### Anakin_k

y=2x2-2x+3

When you complete the square, you must factor out the leading co-efficient, which in this case is 2 from the terms with the variable. The 3 however will stay outside of the brackets.

y=2(x2-x)+3
y=2(x2-x+0.25-0.25)+3
Bring out the negative 0.25 after multiplying it by the leading co-efficient that was factored out but leave the positive 0.25 inside the brackets. Now simply.

You should be able to take it from there.