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Graph x^2 - y^2 = 1 from a parametric definition. Part of HELM Book 2.2.2.

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Consider the function $x=\\dfrac{1}{2} \\left( t + \\dfrac{1}{t} \\right), \\; y = \\dfrac{1}{2} \\left( t-\\dfrac{1}{t} \\right), \\; 1 \\leq t \\leq 8$.

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$t$$1$$2$$3$$4$$5$$6$$7$$8$
$x$$1$$1.25$$1.67$$2.43$$2.60$$3.08$$3.57$$4.06$
$y$$0$$0.75$$1.33$$1.88$$2.40$$2.92$$3.43$$3.94$
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{graph}

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(a) Draw up a table of values of this function. A partially completed table of values has been prepared. Complete the table.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$t$$1$$2$$3$$4$$5$$6$$7$$8$
$x$$1$$1.25$$1.67$[[0]][[1]][[2]][[3]]$4.06$
$y$$0$$0.75$[[4]][[5]][[6]][[7]][[8]]$3.94$
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(b) Plot a graph of the function. A graph is shown in the figure. Add your points to those already marked on the graph.

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It is possible to eliminate t between the two equations so that the original parametric form can be expressed as $x^2 − y^2 = 1$.

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