Question Number 29960 by rahul 19 last updated on 14/Feb/18
Answered by ajfour last updated on 15/Feb/18
![(x^2 /a^2 )+(y^2 /b^2 )=1 ⇒ (x/a^2 )+(y/b^2 )((dy/dx))=0 slope of a normal = −(dx/dy) = n n= (a^2 /b^2 )((y/x)) and if we let x=acos θ , y=bsin θ then n= ((asin θ)/(bcos θ)) eq. of a normal through (h, k) is y−k = ((asin θ)/(bcos θ))(x−h) ⇒ (bsin θ−k)bcos θ=asin θ(acos θ−h) −kbcos θ=sin θ(a^2 cos θ−b^2 cos θ−ah) k^2 b^2 cos^2 θ=(1−cos^2 θ)[(a^2 −b^2 )cos θ−ah]^2 let us say cos θ = t k^2 b^2 t^2 =(1−t^2 )[(a^2 −b^2 )^2 t^2 +a^2 h^2 −2ah(a^2 −b^2 )t ] k^2 b^2 t^2 =(a^2 −b^2 )^2 t^2 +a^2 h^2 −2ah(a^2 −b^2 )t −(a^2 −b^2 )^2 t^4 −a^2 h^2 t^2 +2ah(a^2 −b^2 )t^3 finally (a^2 −b^2 )^2 t^4 −2ah(a^2 −b^2 )t^3 + kt^2 +2ah(a^2 −b^2 )t−a^2 h^2 =0 Σcos α =Σt_1 = ((2ah)/((a^2 −b^2 ))) Σsec α =((Σcos αcos βcos γ)/(Πcos α)) = ((Σt_1 t_2 t_3 )/(Πt_1 )) = (((((2ah)/(a^2 −b^2 ))))/([((a^2 h^2 )/((a^2 −b^2 )^2 ))])) =((2(a^2 −b^2 ))/(ah)) ⇒ (Σcos α)(Σsec α) = ((2ah)/((a^2 −b^2 )))×((2(a^2 −b^2 ))/(ah)) = 4 .](https://www.tinkutara.com/question/Q30053.png)
$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\Rightarrow\:\:\frac{{x}}{{a}^{\mathrm{2}} }+\frac{{y}}{{b}^{\mathrm{2}} }\left(\frac{{dy}}{{dx}}\right)=\mathrm{0} \\ $$$${slope}\:{of}\:{a}\:{normal}\:=\:−\frac{{dx}}{{dy}}\:=\:{n} \\ $$$${n}=\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\left(\frac{{y}}{{x}}\right)\:\:{and}\:{if}\:{we}\:{let} \\ $$$${x}={a}\mathrm{cos}\:\theta\:\:,\:{y}={b}\mathrm{sin}\:\theta\:\:{then} \\ $$$${n}=\:\frac{{a}\mathrm{sin}\:\theta}{{b}\mathrm{cos}\:\theta} \\ $$$${eq}.\:{of}\:{a}\:{normal}\:{through}\:\left({h},\:{k}\right)\:{is} \\ $$$${y}−{k}\:=\:\frac{{a}\mathrm{sin}\:\theta}{{b}\mathrm{cos}\:\theta}\left({x}−{h}\right) \\ $$$$\Rightarrow\:\left({b}\mathrm{sin}\:\theta−{k}\right){b}\mathrm{cos}\:\theta={a}\mathrm{sin}\:\theta\left({a}\mathrm{cos}\:\theta−{h}\right) \\ $$$$−{kb}\mathrm{cos}\:\theta=\mathrm{sin}\:\theta\left({a}^{\mathrm{2}} \mathrm{cos}\:\theta−{b}^{\mathrm{2}} \mathrm{cos}\:\theta−{ah}\right) \\ $$$${k}^{\mathrm{2}} {b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta=\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta\right)\left[\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{cos}\:\theta−{ah}\right]^{\mathrm{2}} \\ $$$${let}\:{us}\:{say}\:\mathrm{cos}\:\theta\:=\:{t} \\ $$$${k}^{\mathrm{2}} {b}^{\mathrm{2}} \boldsymbol{{t}}^{\mathrm{2}} =\left(\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} \right)\left[\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \boldsymbol{{t}}^{\mathrm{2}} +{a}^{\mathrm{2}} {h}^{\mathrm{2}} \right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{ah}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\boldsymbol{{t}}\:\right] \\ $$$${k}^{\mathrm{2}} {b}^{\mathrm{2}} \boldsymbol{{t}}^{\mathrm{2}} =\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \boldsymbol{{t}}^{\mathrm{2}} +{a}^{\mathrm{2}} {h}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{ah}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\boldsymbol{{t}}\:−\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \boldsymbol{{t}}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−{a}^{\mathrm{2}} {h}^{\mathrm{2}} \boldsymbol{{t}}^{\mathrm{2}} +\mathrm{2}{ah}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\boldsymbol{{t}}^{\mathrm{3}} \\ $$$${finally} \\ $$$$\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \right)^{\mathrm{2}} \boldsymbol{{t}}^{\mathrm{4}} −\mathrm{2}\boldsymbol{{ah}}\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \right)\boldsymbol{{t}}^{\mathrm{3}} \\ $$$$\:\:+\:\boldsymbol{{kt}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{ah}}\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \right)\boldsymbol{{t}}−\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{h}}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Sigma\mathrm{cos}\:\alpha\:=\Sigma{t}_{\mathrm{1}} =\:\frac{\mathrm{2}\boldsymbol{{ah}}}{\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \right)} \\ $$$$\Sigma\mathrm{sec}\:\alpha\:=\frac{\Sigma\mathrm{cos}\:\alpha\mathrm{cos}\:\beta\mathrm{cos}\:\gamma}{\Pi\mathrm{cos}\:\alpha} \\ $$$$\:=\:\frac{\Sigma{t}_{\mathrm{1}} {t}_{\mathrm{2}} {t}_{\mathrm{3}} }{\Pi{t}_{\mathrm{1}} }\:=\:\frac{\left(\frac{\mathrm{2}\boldsymbol{{ah}}}{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} }\right)}{\left[\frac{\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{h}}^{\mathrm{2}} }{\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \right)^{\mathrm{2}} }\right]}\:=\frac{\mathrm{2}\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \right)}{\boldsymbol{{ah}}} \\ $$$$\Rightarrow\:\:\:\left(\Sigma\mathrm{cos}\:\alpha\right)\left(\Sigma\mathrm{sec}\:\alpha\right)\:= \\ $$$$\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}\boldsymbol{{ah}}}{\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \right)}×\frac{\mathrm{2}\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \right)}{\boldsymbol{{ah}}}\:=\:\mathrm{4}\:. \\ $$$$ \\ $$
Commented by rahul 19 last updated on 15/Feb/18
$$\mathrm{thank}\:\mathrm{u}\:\mathrm{sir}! \\ $$