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Question Number 45638 by peter frank last updated on 14/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(asec\theta ,btan\theta )liesonhyperbola$$$$eqnoftangentat(x_1,y_1)is$$$$\frac{{xx}_1}{a^2}-\frac{{yy}_1}{b^2}=1here(x_1,y_1)=(asec\theta ,btan\theta )$$$$\frac{x\times asec\theta }{a^2}-\frac{y\times btan\theta }{b^2}=1$$$$\frac{xsec\theta }{a}-\frac{ytan\theta }{b}=1$$$$thefociofhyperbolaare(\pm \sqrt{a^2+b^2},0)$$$$xbsec\theta -yatan\theta -ab=0$$$$distancdfrom\left(\sqrt{a^2+b^2}\overline{,}0\right)totheabovetangent$$$$is{P}_{1}and(-\sqrt{a^2+b^2},0){isP}_2$$$$toprove\mid P_1{P}_2\mid =b^2$$$$P_1=\mid \frac{\left(\sqrt{a^2+b^2}\times bsec\theta \right)-ab}{\sqrt{b^2{sec}^2\theta +a^2{tan}^2\theta }}\mid$$$$P_2=\mid \frac{-\sqrt{a^2+b^2}\times bsec\theta -ab}{\sqrt{b^2{sec}^2\theta +a^2{tan}^2\theta }}\mid$$$$P_1{P}_2=\frac{(a^2+b^2)b^2{sec}^2\theta -a^2{b}^2}{b^2{sec}^2\theta +a^2{tan}^2\theta }$$$$=b^2\times \frac{(a^2+b^2){sec}^2\theta -a^2}{b^2{sec}^2\theta +a^2({sec}^2\theta -1)}$$$$=b^2\times \frac{(a^2+b^2){sec}^2\theta -a^2}{(a^2+b^2){sec}^2\theta -a^2}=b^{2}proved$$$$\stackrel{=}{}$$$$$$

Commented by peter frank last updated on 20/Oct/18

$$\mathrm{sorry}\mathrm{sir}\mathrm{i}\mathrm{dont}\mathrm{it}\mathrm{get}\mathrm{where}\mathrm{this}\mathrm{come}\mathrm{from}(\pm \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2},0)$$

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