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P5 watch

1. Can someone just check that I haven't fudged my way to the correct answer?

Find in cartesian form the equation of the parabola whose focus is (3,0) and directrix is x+5=0?

because of the focus-directrix property (0,-1) is a point on the parabola.

x=2at now goes to (x+1)=4(t^2) and y=8t

subbing in for y gives:

(x+1)=4(y^2/64)
so y^2 =16x+16
2. I can't quite follow what you've done. But here's how I'd do it:

Suppose (x,y) is on the parabola. Then we can use the focus-directrix property:
sqrt((x-3)^2 + y^2) = sqrt((x+5)^2 + (y-y)^2)
x^2 - 6x + 9 + y^2 = x^2 + 10x + 25
=> y^2 = 16x + 16
3. I can't quite follow what you've done. But here's how I'd do it
I think its safe to say I fudged it. It didn't work with the next question. You method looks sweeter even if mine was right. Thankyou!
4. (Original post by ssmoose)
Can someone just check that I haven't fudged my way to the correct answer?

Find in cartesian form the equation of the parabola whose focus is (3,0) and directrix is x+5=0?

because of the focus-directrix property (0,-1) is a point on the parabola.

x=2at now goes to (x+1)=4(t^2) and y=8t

subbing in for y gives:

(x+1)=4(y^2/64)
so y^2 =16x+16
Hmm, could you explain what you've done?
dvs' method is the one I use.
5. See above, lol. I think I was away when this stuff was taught and never caught it up. Nice to see someone else going to Durham by the way Is there a P5/FP2 revision thread anywhere?
6. (Original post by ssmoose)
Is there a P5/FP2 revision thread anywhere?
7. it's closed. Is it possible if I ask a moderator to have it reopened?
8. (Original post by ssmoose)
See above, lol. I think I was away when this stuff was taught and never caught it up. Nice to see someone else going to Durham by the way Is there a P5/FP2 revision thread anywhere?
Hmm, not that I've seen.
There might be some from last year though.
edit: ah it's been found.
9. (Original post by ssmoose)
it's closed. Is it possible if I ask a moderator to have it reopened?
Yeah. Just go to the first post of the thread, for example, report it (red triangle) and in the report ask for it to be reopened.
10. How would we go about doing the opposite then?

For example how would we find the focus and drectrix of y^2=4x+4
11. y^2 = 4(x+1)

Compare this to y^2 = 4ax, whose focus is at (a, 0) and directrix at x=-a. We see that a=1, so this would imply that (1, 0) and x=-1 are the focus and directrix. However we also have a translation of 1 unit to the left. So the focus and directrix actually are (0, 0) and x=-2.
12. (Original post by ssmoose)
How would we go about doing the opposite then?

For example how would we find the focus and drectrix of y^2=4x+4
y^2 = 4aX = 4(x+1)
=> a = 1
=> X = x+1

Directrix: X+a = 0 => x+1+1 = 0 => x+2=0
Focus: X = a, y = 0 => x+1=1, y = 0 => x = 0, y = 0
13. arghh why is tsr so slow today?

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