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Question Number 68783 by peter frank last updated on 15/Sep/19

Commented by peter frank last updated on 22/Sep/19

$$thankyou$$

Commented by ajfour last updated on 15/Sep/19

Commented by peter frank last updated on 21/Sep/19

$$pleasesir.continue$$

Commented by ajfour last updated on 21/Sep/19

$$letS{}^{\prime }P=a+s,SP=a-s$$$$letP(h,k)$$$$\frac{S{}^{\prime }P}{SP}=\frac{a+s}{a-s}$$$$a-s=e(\frac{a}{e}-h)$$$$a+s=e(\frac{a}{e}+h)$$$$\Rightarrow s=eh$$$$\frac{S{}^{\prime }P}{SP}=\frac{a+eh}{a-eh}................\left(i\right)$$$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow -\frac{dx}{dy}=\frac{a^2}{b^2}\left(\frac{y}{x}\right)$$$$slopeofnormal$$$$=-\frac{dx}{dy}{\mid }_P=\frac{a^2}{b^2}\left(\frac{k}{h}\right)$$$$eq.ofnormal$$$$y-k=\frac{a^2}{b^2}\left(\frac{k}{h}\right)(x-h)$$$$letnormalintersectx-axis$$$$atN(0,x_{int})$$$$x_{int}=h(1-\frac{b^2}{a^2})$$$$\frac{SN}{S{}^{\prime }N}=\frac{ae-x_{int}}{x_{int}+ae}=\frac{ae-h(1-\frac{b^2}{a^2})}{h(1-\frac{b^2}{a^2})+ae}=r_2$$$$\Rightarrow \frac{SN}{S{}^{\prime }N}=\frac{a-eh}{eh+a}.............\left(ii\right)$$$$from\left(i\right)and\left(ii\right)wefind$$$$\frac{S{}^{\prime }P}{SP}=\frac{S{}^{\prime }N}{SN}$$$$\Rightarrow PNisangularbisector$$$$of\mathrm{\angle }S{}^{\prime }PS.$$

Commented by peter frank last updated on 22/Sep/19

$$pleasehelp69230$$

Answered by mr W last updated on 15/Sep/19

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