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1. C4

I know you have to do more work to find the actual value of a, but it says a(2t-1)=0, so isn't a=0 a solution?
2. Re: C4
(Original post by xXxiKillxXx)

I know you have to do more work to find the actual value of a, but it says a(2t-1)=0, so isn't a=0 a solution?
You are confusing constants with variables.

If you have then

But if you have then and not

is a constant.
3. Re: C4
It doesn't satisfy the x and y equations, only the y.
4. Re: C4
(Original post by raheem94)
You are confusing constants with variables.

If you have then

But if you have then and not

is a constant.
Something can only be a constant if it has one value right? So Isn't x a constant also?
5. Re: C4
(Original post by xXxiKillxXx)
Something can only be a constant if it has one value right? So Isn't x a constant also?
Yes, constant has only one value.

is a variable, it can take many values.

In the question it is specified that is a constant.
6. Re: C4
(Original post by raheem94)
Yes, constant has only one value.

is a variable, it can take many values.

In the question it is specified that is a constant.
Ah thanks
7. Re: C4
(Original post by raheem94)
Yes, constant has only one value.

is a variable, it can take many values.

In the question it is specified that is a constant.
Wait, if a is a constant and it can only take one value, how do we know (initially) that value isn't 0 as opposed to 2x-1 = a ?
8. Re: C4
(Original post by xXxiKillxXx)
Wait, if a is a constant and it can only take one value, how do we know (initially) that value isn't 0 as opposed to 2x-1 = a ?
If a=0 then , then will be zero for all values of , so how can the curve pass through the point
9. Re: C4
(Original post by raheem94)
If a=0 then , then will be zero for all values of , so how can the curve pass through the point
That makes sense, thanks!
10. Re: C4
x=4=4at² --> rearranged to at²=1
y=0=a(2t+1) --> rearranged to t=-0.5 (2at+a=0>>2at=-a>>2t=-1>>t=-0.5)

put t=-0.5 into at²=1 then you get a= 1/t² (1/0.25)

therefore a=4

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