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 50676 by Tinkutara last updated on 18/Dec/18

Commented by ajfour last updated on 18/Dec/18

Commented by ajfour last updated on 18/Dec/18

E((a/2)sin^2 θ, (a/2)cos θ sin θ) m_(BE) = ((sin θcos θ)/(1+sin^2 θ)) m_(AF) = −((tan θ−((sin θcos θ)/2))/((sin^2 θ)/2)) = − ((2−cos^2 θ)/(sin θcos θ)) = −(1/m_(BE) ) hence AF ⊥ BE .

$${E}\left(\frac{{a}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \theta,\:\frac{{a}}{\mathrm{2}}\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\right) \\ $$$${m}_{{BE}} \:=\:\frac{\mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \theta} \\ $$$${m}_{{AF}} \:=\:−\frac{\mathrm{tan}\:\theta−\frac{\mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{2}}}{\frac{\mathrm{sin}\:^{\mathrm{2}} \theta}{\mathrm{2}}} \\ $$$$\:\:\:\:\:=\:−\:\frac{\mathrm{2}−\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta\mathrm{cos}\:\theta}\:=\:−\frac{\mathrm{1}}{{m}_{{BE}} } \\ $$$${hence}\:\:\:{AF}\:\bot\:{BE}\:. \\ $$

Commented by Tinkutara last updated on 19/Dec/18

Thanks Sir! Excellent method, simple and short!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com