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# FP1 Watch

1. How do you work out the general equation of the normal to P(at^2, at) on the curve y^2=4ax

if you differentiate to get 2a/y as the gradient
2. (Original post by Acrux)
How do you work out the general equation of the normal to P(at^2, at) on the curve y^2=4ax

if you differentiate to get 2a/y as the gradient
Is the bit I've bolded a typo? I'm assuming it is.

y-2at = -t(x-at^2) and simplify. Ta-da.
3. (Original post by Zacken)
Is the bit I've bolded a typo? I'm assuming it is.

y-2at = -t(x-at^2) and simplify. Ta-da.
Yes it was.
So differentiating y^2=4ax gives 2a/y why put the y-value and not x?
I dont understand this part
4. (Original post by Acrux)
Yes it was.
So differentiating y^2=4ax gives 2a/y why put the y-value and not x?
...because it's 2a/y. Had it been 2a/x then you'd have put the x value in.
5. (Original post by Zacken)
...because it's 2a/y. Had it been 2a/x then you'd have put the x value in.
So this doesn't matter just depends on which it is y or x
6. (Original post by Acrux)
So this doesn't matter just depends on which it is y or x
What? If you have a y in your gradient function, then replace the y with the y coordinate to find the gradient at that point. If you have an x in your gradient function then replace the x with the x coordinate to find the gradient at that point. If you have both x and y in your gradient function, then replace their them with their suitable coordinates in their respective places.

It's like saying y = 2x^2, to find the y value at x=0, you plug x=0 wherever you see x. So y = 2(0^2) = 0.

Which bit is confusing you?
7. (Original post by Zacken)
What? If you have a y in your gradient function, then replace the y with the y coordinate to find the gradient at that point. If you have an x in your gradient function then replace the x with the x coordinate to find the gradient at that point. If you have both x and y in your gradient function, then replace their them with their suitable coordinates in their respective places.

It's like saying y = 2x^2, to find the y value at x=0, you plug x=0 wherever you see x. So y = 2(0^2) = 0.

Which bit is confusing you?
so it depends on gradient function
thanks
8. (Original post by Acrux)
so it depends on gradient function
thanks
Well, if you say so, I guess.

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