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Question Number 50754 by peter frank last updated on 19/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Dec/18

$$R(asec\theta ,btan\theta )centre(0,0)S(c,0)S^{{}^{\prime }}(-c,0)c=\sqrt{a^2+b^2}$$$$tangent\frac{xsec\theta }{a}-\frac{ytan\theta }{b}=1$$$$distancefrom(0,0)isp$$$$p=\mid \frac{-1}{\sqrt{{\left(\frac{sec\theta }{a}\right)}^2+{(-\frac{tan\theta }{b})}^2}}\mid$$$$\frac{1}{p^2}=\frac{{sec}^2\theta }{a^2}+\frac{{tan}^2\theta }{b^2}$$$$LHS$$$$4a^2(1+\frac{b^2}{p^2})$$$$=4a^2+\frac{4a^2{b}^2}{p^2}$$$$=4a^2+4a^2{b}^2(\frac{{sec}^2\theta }{a^2}+\frac{{tan}^2\theta }{b^2})$$$$=4a^2+4b^2{sec}^2\theta +4a^2{tan}^2\theta$$$$=4a^2(1+{tan}^2\theta )+4b^2{sec}^2\theta$$$$=4{sec}^2\theta \times (a^2+b^2)$$$$=4c^2{sec}^2\theta [since{a}^2+b^2=c^2]$$$$RHS$$$${(RS+{RS}^{\prime })}^2$$$$RS=\sqrt{{(asec\theta -c)}^2+{\left(btan\theta \right)}^2}$$$$=\sqrt{a^2{sec}^2\theta -2acsec\theta +c^2+b^2({sec}^2\theta -1)}$$$$=\sqrt{(a^2+b^2){sec}^2\theta -2acsec\theta +c^2-b^2}$$$$=\sqrt{c^2{sec}^2\theta -2acsec\theta +a^2}$$$$=csec\theta -a$$$${RS}^{{}^{\prime }}=csec\theta +a$$$${(RS+{RS}^{{}^{\prime }})}^2$$$$={\left(2csec\theta \right)}^2=4c^2{sec}^2\theta$$$$proved$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$R$$

Answered by ajfour last updated on 20/Dec/18

$$S(ae,0);S_1(-ae,0)$$$$R(a\sec \theta ,b\tan \theta )$$$$\frac{dy}{dx}=\frac{b}{a\sin \theta }$$$$eq.oftangent$$$$\frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1$$$$x_{intercept}=a\cos \theta$$$$Also{x}_{intercept}=\frac{p}{\sin \theta }$$$$\Rightarrow p=a\sin \theta \cos \theta$$$$RS=e(x_R-\frac{a}{e}),{RS}_1=e(x_R+\frac{a}{e})$$$$\Rightarrow {(RS+{RS}_1)}^2=4e^2{x}_R^2$$$$=4a^2{e}^2\sec {}^2\theta$$$$andwith{e}^2=1+\frac{b^2}{a^2}$$$$idontgetr.h.s.$$$$pleasehelptroubleshoot..$$

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