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Given the base of a triangle and the ratio of the tangent of half the base angles.The vertex moves on a :

 

Option: 1

Hyperbola
 


Option: 2

 Ellipse 


Option: 3

circle
 


Option: 4

Parabola


Answers (1)

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\mathrm{\mathrm{BC}= base\, \, of\, \, the\, \, triangle =\mathrm{a} (constant) }and A is the vertex.
\mathrm{\begin{aligned} & \frac{\tan B / 2}{\tan C / 2}=\frac{\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}}{\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}} \\ & =\frac{s-c}{s-b} \\ & =\frac{2 s-2 c}{2 s-2 b} \\ & =\frac{a+b-c}{a-b+c}=k \text { (let) } \end{aligned} }

By componendo and dividendo
\mathrm{\begin{array}{ll} \Rightarrow & \frac{\mathrm{k}-1}{\mathrm{k}+1}=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{a}} \\ \Rightarrow & \mathrm{b}-\mathrm{c}=\mathrm{a}\left(\frac{\mathrm{k}-1}{\mathrm{k}+1}\right)=\text { constant } \\ \Rightarrow & \mathrm{CA}-\mathrm{BA}=\text { constant } \end{array} }
By the focal property, the locus of A is a hyperbola whose foci are B & C

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Gaurav

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