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Worked solution

Eliminate reciprocal parameters and retain both hyperbola branches

For x=t+1/tx=t+1/t, y=t−1/ty=t-1/t, find a Cartesian equation and sketch the curve, showing axial intercepts and asymptotes.

[N2010/I/11(iii)]

CJC · Lecture · Example 6

Use the source sum-and-substitution method

Adding gives x+y=2tx+y=2t, so t=(x+y)/2t=(x+y)/2. Substituting into x gives x=(x+y)/2+2/(x+y)=[(x+y)2+4]/[2(x+y)]x=(x+y)/2+2/(x+y)=[(x+y)^2+4]/[2(x+y)]. Therefore 2x2+2xy=x2+2xy+y2+42x^2+2xy=x^2+2xy+y^2+4, so x2−y2=4x^2-y^2=4, equivalently x2/4−y2/4=1x^2/4-y^2/4=1. The original equations require t≠0t\ne0; on this hyperbola x+y≠0x+y\ne0, so every real point of both branches has an admissible t.

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Identify intercepts and asymptotes

The hyperbola is centred at (0,0)(0,0), with x-intercepts (−2,0)(-2,0) and (2,0)(2,0), no y-intercept, and asymptotes y=xy=x and y=−xy=-x. The source obtains the asymptotes from x2/4−y2/4=0x^2/4-y^2/4=0. Positive t traces the right branch; negative t traces the left branch.

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Compare your sketch

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Both branches of x²−y²=4, centre O, exact intercepts (−2,0),(2,0) and dashed asymptotes y=x and y=−x.
Both branches of x²−y²=4, centre O, exact intercepts (−2,0),(2,0) and dashed asymptotes y=x and y=−x.
Both branches of x²−y²=4, centre O, exact intercepts (−2,0),(2,0) and dashed asymptotes y=x and y=−x.

Common mistakes

Do not retain a single branch merely because a calculator uses a positive minimum t.

Read the hint