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A curve is defined by the parametric equations $x=\\var{a}t+\\frac{\\var{b}}{t^2}, y = \\var{a}t-\\frac{\\var{b}}{t^2}$

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(i) To find $P$ we substitute $t=\\var{t}$ into $x=\\var{a}t+\\frac{\\var{b}}{t^2}, y = \\var{a}t-\\frac{\\var{b}}{t^2}$

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Hence $x=\\var{a}\\times\\var{t}+\\frac{\\var{b}}{\\var{t}^2} = \\var{a*t} + \\var{b/t^2}=\\var{x}$

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and $y=\\var{a}\\times\\var{t}-\\frac{\\var{b}}{\\var{t}^2} = \\var{a*t} - \\var{b/t^2}=\\var{y}$

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Hence $P = (\\var{x},\\var{y})$

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(ii) To find the equation of the tangent, we need to find the gradient $m$, which is simply the value of $dy/dx$ at the point $P$. Since we are working with parametric equations, it is easiest to calculate $dy/dx=\\frac{dy}{dt} / \\frac{dx}{dt}$, and evaluate these when $t = \\var{t}$, corresponding to the point $P$

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Now $\\frac{dy}{dt} = \\var{a}-\\frac{\\var{b}}{t^3} = \\var{a}-\\frac{\\var{b}}{\\var{t}^3} =\\var{ydot}$ at the point $P$

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and $\\frac{dx}{dt} = \\var{a}+\\frac{\\var{b}}{t^3} = \\var{a}+\\frac{\\var{b}}{\\var{t}^3}= \\var{xdot}$ at the point $P$

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Hence $m=dy/dx = \\frac{\\var{ydot}}{\\var{xdot}}=\\simplify[all,fractionNumbers]{{m}}$

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Now we can find $c$ by substituting $m=\\simplify[all,fractionNumbers]{{m}}, x=\\var{x}$ and $y=\\var{y}$ into $y = mx+c$

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Hence $c=y-mx=\\var{y}-\\simplify[all,fractionNumbers]{{m}}\\times \\var{x} = \\simplify[all,fractionNumbers]{{c}}$

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(i) At the point $P$ on the curve, $t=\\var{t}$. Find the co-ordinates of $P$.

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$P$ = ([[0]],[[1]])

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(ii) Find the equation of the tangenet to the curve at point $P$, in the form $y = mx+c$.

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$m$ = [[2]] (give your answer as a fraction)

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$c$ = [[3]] 

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