A (-3,4) and B(2,-1) be two fixed points. A point C moves such that tan⁡12∠ABC:tan⁡12∠BAC=3:1. If the locus of ‘ C ‘ is a hyperbola, then the distancebetween two foci is

tan⁡B2tan⁡A2=3⇒Δs(s−b)Δs(s−a)=3⇒s−as−b=3⇒s−a=3s−3b⇒2s=3b−2a⇒a+b+c=3b−a⇒b−a=c2⇒AC−BC= Constant Distance between two foci =|AB|=52

A (-3,4) and B(2,-1) be two fixed points. A point C moves such that tan⁡12∠ABC:tan⁡12∠BAC=3:1. If the locus of ‘ C ‘ is a hyperbola, then the distancebetween two foci is

tan⁡B2tan⁡A2=3⇒Δs(s−b)Δs(s−a)=3⇒s−as−b=3⇒s−a=3s−3b⇒2s=3b−2a⇒a+b+c=3b−a⇒b−a=c2⇒AC−BC= Constant Distance between two foci =|AB|=52

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A (-3,4) and B(2,-1) be two fixed points. A point C moves such that $\tan (\frac{1}{2} ∠ ABC) : \tan (\frac{1}{2} ∠ BAC) = 3 : 1$ . If the locus of ‘ C ‘ is a hyperbola, then the distance
between two foci is

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$5 \sqrt{2}$

$\frac{5}{3}$

$\frac{5}{\sqrt{2}}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{\tan \frac{B}{2}}{\tan \frac{A}{2}} = 3 \Rightarrow \frac{\frac{Δ}{s (s - b)}}{\frac{Δ}{s (s - a)}} = 3 \Rightarrow \frac{s - a}{s - b} = 3 \Rightarrow s - a = 3 s - 3 b \\ \Rightarrow 2 s = 3 b - 2 a \Rightarrow a + b + c = 3 b - a \Rightarrow b - a = \frac{c}{2} \Rightarrow AC - BC = Constant \end{array}$
Distance between two $foci = | AB | = 5 \sqrt{2}$