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 112606 by ajfour last updated on 08/Sep/20

Commented by ajfour last updated on 10/Sep/20

$$FindminimumlengthofAEin$$$$termsofr,a,b.$$

Answered by mr W last updated on 09/Sep/20

Commented by mr W last updated on 09/Sep/20

$$\mu =\frac{b}{a}$$$$\rho =\frac{r}{a}$$$$\eta =\frac{h}{a}$$$$\lambda =\tan \phi$$$$eqn.ofQA:$$$$y=-k+(x+h)\tan \phi$$$$\lambda x-y+(\lambda h-k)=0$$$${\lambda }^2{a}^2+b^2={(\lambda h-k)}^2$$$$\Rightarrow k=\lambda h-\sqrt{{\lambda }^2{a}^2+b^2}$$$$\Rightarrow \frac{k}{a}=\lambda \eta -\sqrt{{\lambda }^2+{\mu }^2}$$$$P(-a\cos \theta ,-b\sin \theta )$$$$\tan \varphi =\frac{\mu }{\tan \theta }$$$$\phi +\alpha +(\frac{\pi }{2}-\phi -\varphi )=\frac{\pi }{2}$$$$\Rightarrow \alpha \Rightarrow \varphi$$$$-a\cos \theta =-h+r\sin \varphi$$$$\Rightarrow \rho \sin \varphi =\eta -\cos \theta$$$$-b\sin \theta =-k+r\cos \varphi$$$$\Rightarrow \rho \cos \varphi =\lambda \eta -\sqrt{{\lambda }^2+{\mu }^2}-\mu \sin \theta$$$$\Rightarrow \tan \varphi =\frac{\eta -\cos \theta }{\lambda \eta -\sqrt{{\lambda }^2+{\mu }^2}-\mu \sin \theta }$$$$\Rightarrow \frac{\mu }{\tan \theta }=\frac{\eta -\cos \theta }{\lambda \eta -\sqrt{{\lambda }^2+{\mu }^2}-\mu \sin \theta }$$$$(\mu \lambda -\tan \theta )\eta =\mu \sqrt{{\lambda }^2+{\mu }^2}-(1-{\mu }^2)\sin \theta$$$$\Rightarrow \eta =\frac{\mu \sqrt{{\lambda }^2+{\mu }^2}-(1-{\mu }^2)\sin \theta }{\mu \lambda -\tan \theta }$$$$\Rightarrow {\rho }^2={(\eta -\cos \theta )}^2+{(\lambda \eta -\sqrt{{\lambda }^2+{\mu }^2}-\mu \sin \theta )}^2$$$$$$$$x_A=-h+r\cos \phi$$$$y_A=-k+r\sin \phi$$$${AE}^2={(h-r\cos \phi )}^2+{(k-r\sin \phi )}^2$$$$\mathrm{\Phi }={\left(\frac{AE}{a}\right)}^2={(\eta -\rho \cos \phi )}^2+{(\lambda \eta -\sqrt{{\lambda }^2+{\mu }^2}-\rho \sin \phi )}^2$$$$\mathrm{\Phi }={[\frac{\mu \sqrt{{\lambda }^2+{\mu }^2}-(1-{\mu }^2)\sin \theta }{\mu \lambda -\tan \theta }-\rho \cos \phi ]}^2+{[\frac{\mu \lambda \sqrt{{\lambda }^2+{\mu }^2}-\lambda (1-{\mu }^2)\sin \theta }{\mu \lambda -\tan \theta }-\sqrt{{\lambda }^2+{\mu }^2}-\rho \sin \phi ]}^2$$$$.....$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com