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Worked solution

Recover a hyperbola from tangent-intercept midpoints

The curve CC has x=3/tx=3/t, y=2ty=2t.

(i) The line y=2xy=2x cuts CC at AA and BB. Find the exact length of ABAB. marks

(ii) The tangent at P=(3/p,2p)P=(3/p,2p) meets the x-axis at DD and the y-axis at EE. Let FF be the midpoint of DEDE. Find the cartesian equation of the curve traced by FF as pp varies. marks

[2017/II/1]

CJC · Review problems · Question 11

Answer

AB=215AB=2\sqrt{15} units; midpoint locus xy=6xy=6.

(i) Solve both intersections and use their distance

2t=2(3/t)2t=2(3/t) gives t2=3t^2=3, so A=(3,23)A=(\sqrt3,2\sqrt3) and B=(−3,−23)B=(-\sqrt3,-2\sqrt3). Then AB=(23)2+(43)2=60=215AB=\sqrt{(2\sqrt3)^2+(4\sqrt3)^2}=\sqrt{60}=2\sqrt{15} units.

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(ii) Find the tangent and its two intercepts

For p≠0p\ne0, dy/dx=2/(−3/t2)=−2t2/3dy/dx=2/(-3/t^2)=-2t^2/3. The tangent at PP is y−2p=−23p2(x−3/p)y-2p=-\tfrac23p^2(x-3/p), or y=4p−23p2xy=4p-\tfrac23p^2x. Setting y=0y=0 gives D=(6/p,0)D=(6/p,0); setting x=0x=0 gives E=(0,4p)E=(0,4p).

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(ii) Take the midpoint before eliminating the parameter

F=((6/p+0)/2,(0+4p)/2)=(3/p,2p)F=((6/p+0)/2,(0+4p)/2)=(3/p,2p). Thus xy=6xy=6. Every real nonzero pp is allowed, so both branches of the hyperbola occur; the equation already excludes both axes. In fact F=PF=P for every allowed parameter.

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Common mistakes

The reciprocal parameter is nonzero; the locus contains both branches and no point on an axis.

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