Question Number 51588 by peter frank last updated on 28/Dec/18
$${The}\:{tangent}\:{at}\:{P}\:\:{to}\:{an}\:{ellipse} \\ $$$${meets}\:{directrix}\:{at}\:{Q} \\ $$$${prove}\:{that}\:{the}\:{line} \\ $$$${joining}\:{the}\:{corresponding} \\ $$$${focus}\:{to}\:{P}\:{and}\:{Q}\:{are} \\ $$$${perpendicular} \\ $$
Answered by peter frank last updated on 28/Dec/18
![from tangent aysinθ+bxcos θ−ab=0...(i) P=(acos θ,bsin θ) Q=((a/e),y)⇒directrix S=(ae,0) aysinθ+bxcos θ−ab=0 put x=(a/e) aysinθ+b×(a/e)cos θ−ab=0 y=((−b(cos θ−e))/(esin θ)) Q=((a/e), ((−b(cos θ−e))/(esin θ))) S=(ae,0) slope⇒m_(SQ) =((−b(cos θ−e))/(asin θ(1−e^2 ))) P=(acos θ,bsin θ) S=(ae,0) slops⇒m_(SP) =((bsin θ)/(acos θ−ae)) SQ and SP are perpendicula slope_(SQ ) ×slope_(SP ) =−1 m_(SQ) =[((−b(cos θ−e))/(asin θ(1−e^2 ))) ]×[((bsin θ)/(acos θ−ae))] =−(b^2 /(a^2 (1−e^2 ))) [b^(2 ) =a^2 (1−e^2 ) =−(b^2 /b^2 )=−1 hence shown](Q51602.png)
$${from}\:{tangent} \\ $$$${aysin}\theta+{bx}\mathrm{cos}\:\theta−{ab}=\mathrm{0}...\left({i}\right) \\ $$$${P}=\left({a}\mathrm{cos}\:\theta,{b}\mathrm{sin}\:\theta\right) \\ $$$${Q}=\left(\frac{{a}}{{e}},{y}\right)\Rightarrow{directrix} \\ $$$${S}=\left({ae},\mathrm{0}\right) \\ $$$${aysin}\theta+{bx}\mathrm{cos}\:\theta−{ab}=\mathrm{0} \\ $$$${put}\:{x}=\frac{{a}}{{e}}\:\:\: \\ $$$${aysin}\theta+{b}×\frac{{a}}{{e}}\mathrm{cos}\:\theta−{ab}=\mathrm{0} \\ $$$${y}=\frac{−{b}\left(\mathrm{cos}\:\theta−{e}\right)}{{e}\mathrm{sin}\:\theta} \\ $$$${Q}=\left(\frac{{a}}{{e}},\:\:\:\frac{−{b}\left(\mathrm{cos}\:\theta−{e}\right)}{{e}\mathrm{sin}\:\theta}\right) \\ $$$${S}=\left({ae},\mathrm{0}\right) \\ $$$${slope}\Rightarrow{m}_{{SQ}} =\frac{−{b}\left(\mathrm{cos}\:\theta−{e}\right)}{{a}\mathrm{sin}\:\theta\left(\mathrm{1}−{e}^{\mathrm{2}} \right)} \\ $$$${P}=\left({a}\mathrm{cos}\:\theta,{b}\mathrm{sin}\:\theta\right) \\ $$$${S}=\left({ae},\mathrm{0}\right) \\ $$$${slops}\Rightarrow{m}_{{SP}} =\frac{{b}\mathrm{sin}\:\theta}{{a}\mathrm{cos}\:\theta−{ae}} \\ $$$${SQ}\:\:{and}\:{SP}\:{are}\:{perpendicula} \\ $$$${slope}_{{SQ}\:} ×{slope}_{{SP}\:} =−\mathrm{1} \\ $$$${m}_{{SQ}} =\left[\frac{−{b}\left(\mathrm{cos}\:\theta−{e}\right)}{{a}\mathrm{sin}\:\theta\left(\mathrm{1}−{e}^{\mathrm{2}} \right)}\:\:\right]×\left[\frac{{b}\mathrm{sin}\:\theta}{{a}\mathrm{cos}\:\theta−{ae}}\right] \\ $$$$\:\:\:\:\:=−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)}\:\:\left[{b}^{\mathrm{2}\:} ={a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)\right. \\ $$$$\:\:\:\:\:=−\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=−\mathrm{1} \\ $$$${hence}\:{shown} \\ $$$$ \\ $$