If the base of a triangle and the ratio of tangent of half of base angles are given, then identify the locus of the opposite vertex.

Explanation

In triangle ABC, base BC is given.
i.e., BC = a (constant)
Also, given that
`(tan.(B)/(2))/(tan.(C)/(2))=k("constant")`

`therefore" "(sin.(B)/(2)cos.(C)/(2))/(sin.(C)/(2)cos.(B)/(2))=k`
`rArr" "(sin.(B)/(2)cos.(C)/(2)-sin.(C)/(2)cos.(B)/(2))/(sin.(B)/(2)cos.(C)/(2)+sin.(C)/(2)cos.(B)/(2))=(k-1)/(k+1)`
`rArr" "(sin.(B-C)/(2))/(sin.(B+C)/(2))=(k-1)/(k+1)`
`rArr" "(2sin.(B-C)/(2)cos.(B+C)/(2))/(2sin.(B+C)/(2)cos.(B+C)/(2))=(k-1)/(k+1)`
`rArr" "(sinB-sinC)/(sin(B+C))=(k-1)/(k+1)`
`rArr" "(sinB-sinC)/(sinA)=(k-1)/(k+1)"[As in triangle sin (B + C) sin A]"`
`rArr" "(b-c)/(a)=(k-1)/(k+1)" (using sine rule)"`
`rArr" "b-c=a((k-1)/(k+1))`
`rArr" "AC-AB=a((k-1)/(k+1))" (= constant)"`
Thus, difference of two variable sides AC and AB is constant. Hence, locus of vertex A is hyperbola with B and C as foci.