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    The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
    Find the equation of the tangent at P and the co-ordinates of Q.


    So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)
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    (Original post by Improvement)
    The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
    Find the equation of the tangent at P and the co-ordinates of Q.


    So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)
    the tangent is ok
    post a photo of your workings
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    (Original post by Improvement)
    The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
    Find the equation of the tangent at P and the co-ordinates of Q.


    So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)
    Your tangent equation is correct.

    How are you finding the other point of intersection?

    Edit: beat me to it TeeEm, did you do your ... trick
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    (Original post by Improvement)
    The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
    Find the equation of the tangent at P and the co-ordinates of Q.


    So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)
    [I'm assuming you are asking how to get the coordinates of Q.]

    You have the tangent correct. The tangent intersects your curve at the point Q, so you are looking for when the curve intersects the tangent.

    Because your tangent has equation y= -2x +3, you can put it equal to y=t^2.

    You can also use the equation x=1/t to find t in terms of x; and substitute accordingly, so you have an equation entirely populated by Xs. Work out the value of x this way, and then find y by substituting into your tangent equation.

    I hope my explanation made sense... I'm new to this.
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    (Original post by Improvement)
    The tangent at point P(1,1) to the curve x= 1/t, y=t^2 intersects the curve at point Q.
    Find the equation of the tangent at P and the co-ordinates of Q.


    So using dy/dx = dy/dt divided by dx/dt, I have got the gradient (m) as -2 then subbing in x=1 and y =1 I have got the equation y = -2x + 3. But I do not think this is right as you do not get the co-ordinates of Q which are (-1/2, 4)
    The tangent is right and from that you can get the point Q

    as x=1/t and y=t^2 so y=1/x^2

    1/x^2=-2x+3 will be a cubic equation but it can be factorized

    -2x^3+3x^2-1=0 => -2x^3+2x^2 +x^2 -1=0 => -2x^2(x-1) + (x+1)(x-1)=0
    Taking out the common factor you can solve the equation
    considering that any of the factors may be zero.
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    ...there's four people all saying the exact same thing.
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    No Q = 3
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    (Original post by Zacken)
    ...there's four people all saying the exact same thing.
    Well it's possible we're all wrong :P
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    (Original post by JLegion)
    Well it's possible we're all wrong :P
    What, both Zacken and TeeEm? :toofunny:
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    (Original post by TheOtherSide.)
    What, both Zacken and TeeEm? :toofunny:
    I said possible, not probable :lol:
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    (Original post by JLegion)
    I said possible, not probable :lol:
    Fair enough. I still can't imagine that, but okay
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    Thank you! Most of you were right who said about making 1/x^2 = -2x +3, as you get the right co-ordinates for Q!
 
 
 
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