Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

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

$${thank}\:{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com