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High school mathematics: the problem of straight line passing through hyperbolic focus.
Pass through a and b respectively, so that the vertical line is aligned to the left, the vertical feet are C and D respectively, and pass through b, so that it is perpendicular to AC and E.

Then │AF│=e│AC│, │BF│=e│BD│,

∵ │ AF │ = 5 │ FB │ │ AC │ = 5 │ BD │, │ AE │ = │ AC │-│ BD │ = 4 │ BD │, and

│AB │=│AF │+│BF │= e(│AC │+│BD │) = 6e│BD│

In the right triangle ABE, cos 60 =│AE│AB │= 0.5, that is, │AB│=2│AE│.

∴e=4/3∴6e│bd│=8│bd│