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    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
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    (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.

    So the gradient of your tangent at P is 2a/(2at) = 1/t. Hence the gradient of your normal is -t. Then:

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

    So the gradient of your tangent at P is 2a/(2at) = 1/t. Hence the gradient of your normal is -t. Then:

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