Question Number 46829 by peter frank last updated on 01/Nov/18
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18
![eqn of focal chord is y=tanθ(x−(√(a^2 −b^2 )) ) focaus((√(a^2 −b^2 )) ,0) length of focal chord (√((x_2 −x_2 )^2 +(y_2 −y_1 )^2 )) solve y=tanθ(x−(√(a^2 −b^2 )) )and (x^2 /a^2 )+(y^2 /b^2 )=1 to get (x_1 ,y_1 ) and(x_2 ,y_2 ) y=tanθ(x−p) (x^2 /a^2 )+((tan^2 θ(x−p)^2 )/b^2 )=1 (p=(√(a^2 −b^2 )) ) b^2 x^2 +a^2 tan^2 θ((x^2 −2xp+p^2 )=a^2 b^2 x^2 (b^2 +a^2 tan^2 θ)−x(2p×a^2 tan^2 θ)+a^2 tan^2 θp^2 −a^2 b^2 =0 x_1 +x_2 =((2p×a^2 tan^2 θ)/((b^2 +a^2 tan^2 θ)))←x_1 +x_2 x_1 x_2 =((a^2 p^2 tan^2 θ−a^2 b^2 )/(b^2 +a^2 tan^2 θ))←x_1 x_2 (x_2 −x_1 )^2 =(x_2 +x_1 )^2 −4x_1 x_2 =((4p^2 a^4 tan^4 θ)/((b^2 +a^2 tan^2 θ)^2 ))−((4a^2 p^2 tan^2 θ−4a^2 b^2 )/(b^2 +a^2 tan^2 θ)) =((4p^2 a^4 tan^4 θ−(b^2 +a^2 tan^2 θ)(4a^2 p^2 tan^2 θ−4a^2 b^2 ))/((b^2 +a^2 tan^2 θ)^2 )) =((4p^2 a^4 tan^4 θ−4a^2 b^2 p^2 tan^2 θ+4a^2 b^4 −4a^4 p^2 tan^4 θ+4a^4 b^2 tan^2 θ)/((b^2 +a^2 tan^2 θ)^2 )) =((4a^2 b^4 −4a^2 b^2 p^2 tan^2 θ+4a^4 b^2 tan^2 θ)/((b^2 +a^2 tan^2 θ)^2 ))←(x_2 −x_1 )^2 =((4a^2 b^4 +4a^2 b^2 tan^2 θ(a^2 −p^2 ))/((b^2 +a^2 tan^2 θ)^2 )) =((4a^2 b^4 +4a^2 b^4 tan^2 θ)/D_r ^2 ) [p^2 =a^2 −b^2 ] =((4a^2 b^4 (1+tan^2 θ))/D_r ^2 )=((4a^2 b^4 sec^2 θ)/D_r ^2 ) so (x_2 −x_1 )=((2ab^2 secθ)/D_r ) y=(x−p)tanθ y_2 −y_1 =tanθ(x_2 −p−x_1 +p) y_2 −y_1 =tanθ(x_2 −x_1 ) {(y_2 −y_1 )^2 +(x_2 −x_1 )^2 }^(1/2) =[(x_2 −x_1 )^2 {tan^2 θ+1}]^(1/2) =(x_2 −x_1 )secθ =((2ab^2 sec^2 θ)/D_r )=((2ab^2 )/(cos^2 θ[b^2 +a^2 tan^2 θ])) =((2ab^2 )/(b^2 cos^2 θ+a^2 sin^2 θ)) ultimately got the answer](https://www.tinkutara.com/question/Q46832.png)
$${eqn}\:{of}\:{focal}\:{chord}\:{is} \\ $$$${y}={tan}\theta\left({x}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\right)\:{focaus}\left(\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:,\mathrm{0}\right) \\ $$$${length}\:{of}\:{focal}\:{chord} \\ $$$$\sqrt{\left({x}_{\mathrm{2}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)^{\mathrm{2}} }\: \\ $$$${solve}\:{y}={tan}\theta\left({x}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\right){and}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${to}\:{get}\:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:{and}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right) \\ $$$${y}={tan}\theta\left({x}−{p}\right) \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{tan}^{\mathrm{2}} \theta\left({x}−{p}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:\:\left({p}=\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\right) \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\left(\left({x}^{\mathrm{2}} −\mathrm{2}{xp}+{p}^{\mathrm{2}} \right)={a}^{\mathrm{2}} {b}^{\mathrm{2}} \right. \\ $$$${x}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right)−{x}\left(\mathrm{2}{p}×{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right)+{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta{p}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left.{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\frac{\mathrm{2}{p}×{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta}{\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right.}\right)\leftarrow{x}_{\mathrm{1}} +{x}_{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} =\frac{{a}^{\mathrm{2}} {p}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta−{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta}\leftarrow{x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$$\: \\ $$$$\:\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} =\left({x}_{\mathrm{2}} +{x}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{4}{x}_{\mathrm{1}} {x}_{\mathrm{2}} \:\:\:\:\:\:\:\:\: \\ $$$$\:\:=\frac{\mathrm{4}{p}^{\mathrm{2}} {a}^{\mathrm{4}} {tan}^{\mathrm{4}} \theta}{\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }−\frac{\mathrm{4}{a}^{\mathrm{2}} {p}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta−\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{4}{p}^{\mathrm{2}} {a}^{\mathrm{4}} {tan}^{\mathrm{4}} \theta−\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right)\left(\mathrm{4}{a}^{\mathrm{2}} {p}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta−\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)}{\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{p}^{\mathrm{2}} {a}^{\mathrm{4}} {tan}^{\mathrm{4}} \theta−\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {p}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta+\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{4}} {p}^{\mathrm{2}} {tan}^{\mathrm{4}} \theta+\mathrm{4}{a}^{\mathrm{4}} {b}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta}{\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {p}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta+\mathrm{4}{a}^{\mathrm{4}} {b}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta}{\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\leftarrow\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{4}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\left({a}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)}{\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{4}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{4}} {tan}^{\mathrm{2}} \theta}{{D}_{{r}} ^{\mathrm{2}} }\:\:\:\:\left[{p}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right] \\ $$$$=\frac{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{4}} \left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{{D}_{{r}} ^{\mathrm{2}} }=\frac{\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{4}} {sec}^{\mathrm{2}} \theta}{{D}_{{r}} ^{\mathrm{2}} } \\ $$$${so}\:\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)=\frac{\mathrm{2}{ab}^{\mathrm{2}} {sec}\theta}{{D}_{{r}} } \\ $$$${y}=\left({x}−{p}\right){tan}\theta \\ $$$${y}_{\mathrm{2}} −{y}_{\mathrm{1}} ={tan}\theta\left({x}_{\mathrm{2}} −{p}−{x}_{\mathrm{1}} +{p}\right) \\ $$$${y}_{\mathrm{2}} −{y}_{\mathrm{1}} ={tan}\theta\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right) \\ $$$$\left\{\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} \right\}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\left[\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} \left\{{tan}^{\mathrm{2}} \theta+\mathrm{1}\right\}\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right){sec}\theta \\ $$$$=\frac{\mathrm{2}{ab}^{\mathrm{2}} {sec}^{\mathrm{2}} \theta}{{D}_{{r}} }=\frac{\mathrm{2}{ab}^{\mathrm{2}} }{{cos}^{\mathrm{2}} \theta\left[{b}^{\mathrm{2}} +{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \theta\right]} \\ $$$$=\frac{\mathrm{2}{ab}^{\mathrm{2}} }{{b}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta+{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta} \\ $$$${ultimately}\:{got}\:{the}\:{answer} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by peter frank last updated on 01/Nov/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\mathrm{thank}\:\mathrm{you} \\ $$$$ \\ $$
Commented by peter frank last updated on 01/Nov/18
$$\mathrm{D}_{\mathrm{r}} \:\mathrm{mean}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18
$${D}_{{r}} ={denominator}…{i}\:{have}\:{placed}\:{D}_{{r}} \:{to}\:{mimise}\: \\ $$$${time}\:…{not}\:{writing}\:{the}\:{value}\:{of}\:{D}_{{r}\:\:} {again}\:{ana}\:{again} \\ $$