Question Number 51694 by peter frank last updated on 29/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18
![point p(acosθ,bsinθ) tangent at point p is ((xcosθ)/a)+((ysinθ)/b)=1 ((ysinθ)/b)=1−((xcosθ)/a) y=(b/(sinθ))−(b/(sinθ))×((cosθ)/a)x [slope=−(b/a)cotθ] m_1 ×m_2 =−1 m_2 =((−1)/m_1 )=((−1×a)/(−bcotθ))=(a/b)tanθ so normal at point p is (y−bsinθ)=(a/b)tanθ(x−acosθ) to find point G put y=0 x−acosθ=(b/(atanθ))×−bsinθ x−acosθ=((−b^2 cosθ)/a) x=acosθ−((b^2 cosθ)/a)=((cosθ)/a)(a^2 −b^2 ) point G {((a^2 −b^2 )/a)cosθ,0} to find poit g put x=0 y−bsinθ=((atanθ)/b)(0−acosθ) y−bsinθ=((−a^2 sinθ)/b) y=bsinθ−((a^2 sinθ)/b)=((sinθ)/b)(b^2 −a^2 ) point g ={0,((sinθ)/(b ))(b^2 −a^2 )} CG=((a^2 −b^2 )/a)cosθ cg=((b^2 −a^2 )/b)sinθ a^2 CG^2 +b^2 Cg^2 =(a^2 −b^2 )^2 cos^2 θ+(b^2 −a^2 )^2 sin^2 θ =(a^2 −b^2 )^2 (cos^2 θ+sin^2 θ) =(a^2 −b^2 )^2 p(acosθ,bsinθ) N=(acosθ,0) CN=acosθ CG=((a^2 −b^2 )/a)cosθ ((CG)/(CN))=(((a^2 −b^2 )cosθ)/(a×acosθ))=((a^2 −b^2 )/a^2 ).=(c_F ^2 /a^2 )=((c_F /a))^2 =e^2 [focus(c_(F,) 0) so c_F ^2 .=a^2 −b^2 again=(c_F /a)=e]](Q51695.png)
$${point}\:{p}\left({acos}\theta,{bsin}\theta\right) \\ $$$${tangent}\:{at}\:{point}\:{p}\:\:{is}\:\frac{{xcos}\theta}{{a}}+\frac{{ysin}\theta}{{b}}=\mathrm{1} \\ $$$$\frac{{ysin}\theta}{{b}}=\mathrm{1}−\frac{{xcos}\theta}{{a}} \\ $$$${y}=\frac{{b}}{{sin}\theta}−\frac{{b}}{{sin}\theta}×\frac{{cos}\theta}{{a}}{x}\:\:\left[{slope}=−\frac{{b}}{{a}}{cot}\theta\right] \\ $$$${m}_{\mathrm{1}} ×{m}_{\mathrm{2}} =−\mathrm{1} \\ $$$${m}_{\mathrm{2}} =\frac{−\mathrm{1}}{{m}_{\mathrm{1}} }=\frac{−\mathrm{1}×{a}}{−{bcot}\theta}=\frac{{a}}{{b}}{tan}\theta \\ $$$${so}\:{normal}\:{at}\:{point}\:{p}\:{is} \\ $$$$\left({y}−{bsin}\theta\right)=\frac{{a}}{{b}}{tan}\theta\left({x}−{acos}\theta\right) \\ $$$${to}\:{find}\:{point}\:{G}\:{put}\:{y}=\mathrm{0} \\ $$$${x}−{acos}\theta=\frac{{b}}{{atan}\theta}×−{bsin}\theta \\ $$$${x}−{acos}\theta=\frac{−{b}^{\mathrm{2}} {cos}\theta}{{a}} \\ $$$${x}={acos}\theta−\frac{{b}^{\mathrm{2}} {cos}\theta}{{a}}=\frac{{cos}\theta}{{a}}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$${point}\:{G}\:\left\{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}}{cos}\theta,\mathrm{0}\right\} \\ $$$${to}\:{find}\:{poit}\:{g}\:{put}\:{x}=\mathrm{0} \\ $$$${y}−{bsin}\theta=\frac{{atan}\theta}{{b}}\left(\mathrm{0}−{acos}\theta\right) \\ $$$${y}−{bsin}\theta=\frac{−{a}^{\mathrm{2}} {sin}\theta}{{b}} \\ $$$${y}={bsin}\theta−\frac{{a}^{\mathrm{2}} {sin}\theta}{{b}}=\frac{{sin}\theta}{{b}}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$${point}\:{g}\:=\left\{\mathrm{0},\frac{{sin}\theta}{{b}\:}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\right\} \\ $$$${CG}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}}{cos}\theta \\ $$$${cg}=\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}}{sin}\theta \\ $$$${a}^{\mathrm{2}} {CG}^{\mathrm{2}} +{b}^{\mathrm{2}} {Cg}^{\mathrm{2}} \\ $$$$=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} {cos}^{\mathrm{2}} \theta+\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} {sin}^{\mathrm{2}} \theta \\ $$$$=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \left({cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta\right) \\ $$$$=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$ \\ $$$${p}\left({acos}\theta,{bsin}\theta\right) \\ $$$${N}=\left({acos}\theta,\mathrm{0}\right) \\ $$$${CN}={acos}\theta \\ $$$${CG}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}}{cos}\theta \\ $$$$\frac{{CG}}{{CN}}=\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){cos}\theta}{{a}×{acos}\theta}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }.=\frac{{c}_{{F}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\left(\frac{{c}_{{F}} }{{a}}\right)^{\mathrm{2}} ={e}^{\mathrm{2}} \\ $$$$\left[{focus}\left({c}_{{F},} \mathrm{0}\right)\:\:\:{so}\:{c}_{{F}} ^{\mathrm{2}} .={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right. \\ $$$$\left.{again}=\frac{{c}_{{F}} }{{a}}={e}\right] \\ $$