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Prove-that-the-perpendicilar-tangent-to-the-ellipse-x-2-a-2-y-2-b-2-1-meets-on-the-circle-x-2-y-2-a-2-b-2-

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Question Number 51613 by peter frank last updated on 29/Dec/18
Prove that the perpendicilar tangent to the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 meets on the circle x^2 +y^2 =a^2 +b^2 .
$${Prove}\:{that}\:{the}\:{perpendicilar} \\ $$$${tangent}\:{to}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:{meets}\:{on}\:{the} \\ $$$${circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} . \\ $$
Answered by peter frank last updated on 29/Dec/18
y=mx+(√(a^2 m^2 +b^2 )) y−mx=(√(a^2 m^2 +b^2 )) (y−mx)^2 =((√(a^2 m^2 +b^2 )) )^2 m^2 (x^2 −a^2 )−2mxy+y^2 −b^2 =0 m_(1 ) +m_2 = ((2mx)/(x^2 −a^2 )) m_(1 ) m_2 = ((y^2 −b^2 )/(x^2 −a^2 )) for perpendicular m_1 m_(2 ) =−1 −1= ((y^2 −b^2 )/(x^2 −a^2 )) y^2 +x^2 =a^2 +b^2
$${y}={mx}+\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${y}−{mx}=\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\left({y}−{mx}\right)^{\mathrm{2}} =\left(\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\right)^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} \left({x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)−\mathrm{2}{mxy}+{y}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${m}_{\mathrm{1}\:} +{m}_{\mathrm{2}} =\:\frac{\mathrm{2}{mx}}{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${m}_{\mathrm{1}\:} {m}_{\mathrm{2}} =\:\frac{{y}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${for}\:{perpendicular}\: \\ $$$${m}_{\mathrm{1}} {m}_{\mathrm{2}\:} =−\mathrm{1} \\ $$$$−\mathrm{1}=\:\frac{{y}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} +{x}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$ \\ $$
Answered by mr W last updated on 29/Dec/18
two perpendicular lines from point (h,k): ((y−k)/(x−h))=m ⇒mx−y−mh+k=0 ((y−k)/(x−h))=−(1/m) ⇒x+my−h+mk=0 since they tangent the ellipse, a^2 m^2 +b^2 (−1)^2 −(−mh+k)^2 =0 ⇒a^2 m^2 +b^2 =m^2 h^2 +k^2 −2mhk ...(i) a^2 (1)^2 +b^2 m^2 −(−h+mk)^2 =0 ⇒a^2 +b^2 m^2 =h^2 +m^2 k^2 −2mhk ...(ii) (i)+(ii): ⇒a^2 (m^2 +1)+b^2 (1+m^2 )=h^2 (m^2 +1)+k^2 (1+m^2 ) ⇒a^2 +b^2 =h^2 +k^2 or ⇒x^2 +y^2 =a^2 +b^2
$${two}\:{perpendicular}\:{lines}\:{from}\:{point} \\ $$$$\left({h},{k}\right): \\ $$$$\frac{{y}−{k}}{{x}−{h}}={m}\:\Rightarrow{mx}−{y}−{mh}+{k}=\mathrm{0} \\ $$$$\frac{{y}−{k}}{{x}−{h}}=−\frac{\mathrm{1}}{{m}}\:\Rightarrow{x}+{my}−{h}+{mk}=\mathrm{0} \\ $$$${since}\:{they}\:{tangent}\:{the}\:{ellipse}, \\ $$$${a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \left(−\mathrm{1}\right)^{\mathrm{2}} −\left(−{mh}+{k}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} ={m}^{\mathrm{2}} {h}^{\mathrm{2}} +{k}^{\mathrm{2}} −\mathrm{2}{mhk}\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${a}^{\mathrm{2}} \left(\mathrm{1}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} −\left(−{h}+{mk}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {m}^{\mathrm{2}} ={h}^{\mathrm{2}} +{m}^{\mathrm{2}} {k}^{\mathrm{2}} −\mathrm{2}{mhk}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\Rightarrow{a}^{\mathrm{2}} \left({m}^{\mathrm{2}} +\mathrm{1}\right)+{b}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)={h}^{\mathrm{2}} \left({m}^{\mathrm{2}} +\mathrm{1}\right)+{k}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={h}^{\mathrm{2}} +{k}^{\mathrm{2}} \\ $$$${or} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$
Commented by peter frank last updated on 29/Dec/18
thank you
$${thank}\:{you} \\ $$

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