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    It's part (a).
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    (Original post by Malawi)
    I'm stuck on part a.
    Note that it's just a way of saying that the coordinates are of the form

    x=f(t)

    y=g(t)

    Does that help?
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    (Original post by Indeterminate)
    Note that it's just a way of saying that the coordinates are of the form

    x=f(t)

    y=g(t)

    Does that help?
    No, I've never covered g(y) or y=g(t)
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    (Original post by Malawi)
    No, I've never covered g(y) or y=g(t)
    Have you not covered this in class? This question looks like a piece of standard bookwork to be honest - you should have a chapter in your FP1 book that covers the parametric forms of ellipse, parabola and hyperbola, and various ways of working out the tangent and normal in terms of the parameter t.
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    (Original post by davros)
    Have you not covered this in class? This question looks like a piece of standard bookwork to be honest - you should have a chapter in your FP1 book that covers the parametric forms of ellipse, parabola and hyperbola, and various ways of working out the tangent and normal in terms of the parameter t.
    I've covered this in class and correctly answered many questions similar to this one but I can't seem to do this one. I have tried using the Cartesian equation y^2=4ax and the parametric equations y=2at and x=at^2. I tried subbing in the parametric x equation into the Cartesian equation to get rid of a and then I differentiated and found the gradient and then found the gradient of the normal and put it into (y1-y2)=m(x1-x2). This didn't work so I tried it again but instead I made an equation out of the two parametric equations and then differentiated and so on but this still didn't give me the correct equation.
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    (Original post by Malawi)
    I've covered this in class and correctly answered many questions similar to this one but I can't seem to do this one. I have tried using the Cartesian equation y^2=4ax and the parametric equations y=2at and x=at^2. I tried subbing in the parametric x equation into the Cartesian equation to get rid of a and then I differentiated and found the gradient and then found the gradient of the normal and put it into (y1-y2)=m(x1-x2). This didn't work so I tried it again but instead I made an equation out of the two parametric equations and then differentiated and so on but this still didn't give me the correct equation.
    You do remember parametric differentiation, right?

    Let me give you an example:

    All points on a curve have coordinates

    \left(2t, 3t^2 \right)

    This means that

    x=2t, y=3t^2

    \dfrac{dy}{dx} = \dfrac{6t}{2} = 3t

    This is the same kind of thing!
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    (Original post by Malawi)
    I've covered this in class and correctly answered many questions similar to this one but I can't seem to do this one. I have tried using the Cartesian equation y^2=4ax and the parametric equations y=2at and x=at^2. I tried subbing in the parametric x equation into the Cartesian equation to get rid of a and then I differentiated and found the gradient and then found the gradient of the normal and put it into (y1-y2)=m(x1-x2). This didn't work so I tried it again but instead I made an equation out of the two parametric equations and then differentiated and so on but this still didn't give me the correct equation.
    Have you tried this again using the hint from indeterminate?

    For this example, you know that dy/dt = 9/2 and dx/dt = (9/2)t, so the chain rule tells you that gradient of tangent = dy/dx = (dy/dt) / (dx/dt) = (9/2) / ((9/2)t) = 1/t.

    So for the tangent at V you know that y = (1/t)x + c for some c which you can find by subbing the coordinates of V back into this equation.

    Now follow the same logic for the normal at V (you know how the gradient of a normal relates to the gradient of a tangent) and you should be able to get the answer to part (a).
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    (Original post by davros)
    Have you tried this again using the hint from indeterminate?

    For this example, you know that dy/dt = 9/2 and dx/dt = (9/2)t, so the chain rule tells you that gradient of tangent = dy/dx = (dy/dt) / (dx/dt) = (9/2) / ((9/2)t) = 1/t.

    So for the tangent at V you know that y = (1/t)x + c for some c which you can find by subbing the coordinates of V back into this equation.

    Now follow the same logic for the normal at V (you know how the gradient of a normal relates to the gradient of a tangent) and you should be able to get the answer to part (a).
    No.
 
 
 
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