Subjective Type

Given the base of a triangle and the ratio of the tangent of half the base angles,find the locus of vertex.

Solution

Suppose $$AB$$ to the base and $$C$$ be the moving point,
Then $$A$$ and $$B$$ are the base angle and by hypothesis is
$$\displaystyle \dfrac { \tan { \left( \dfrac { A }{ 2 } \right) } }{ \tan { \left( \dfrac { B }{ 2 } \right) } } =\dfrac { \dfrac { r }{ \left( s-a \right) } }{ \dfrac { r }{ \left( s-b \right) } } =\dfrac { s-b }{ s-a }=$$ constant $$=\lambda $$
$$\displaystyle \Rightarrow \lambda =\dfrac { \dfrac { a+b+c }{ 2 } -a }{ \dfrac { a+b+c }{ 2 } -b } =\dfrac { b+c-a }{ c+a-b } $$
$$\displaystyle \Rightarrow \dfrac { \lambda }{ 1 } =\dfrac { b+c-a }{ c+a-b } $$
Hence, by componendo and dividendo
$$\displaystyle \dfrac { 1+\lambda }{ 1-\lambda } =\dfrac { 2c }{ 2\left( b-a \right) } =\dfrac { c }{ b-a } \Rightarrow b-a=c\dfrac { \left( 1-\lambda \right) }{ \left( 1+\lambda \right) } $$
As $$c$$ is a constant hence, $$(b-a) =$$ difference of distances of a point from two fixed points$$=$$ constant
Therefore, the locus of the vertex $$C$$ is a hyperbola.

SIMILAR QUESTIONS

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