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Question Number 186809 by ajfour last updated on 10/Feb/23

Answered by ajfour last updated on 10/Feb/23

Commented by ajfour last updated on 10/Feb/23

$$x=p=\tan \theta ,let\tan \alpha =c$$$$y=cx$$$$leftcornerA(-p^2,c)$$$$\mathrm{\perp }distanceq$$$$q\sqrt{1+c^2}=c(1+p^2)$$$$letsidesofcentralrectanglebe$$$$sandt.$$$$s=c(1+p^2)-p\sin \alpha$$$$=c(1+p^2)-\frac{cp}{\sqrt{1+c^2}}$$$$t=(1+p^2)\cos \alpha -(c-\frac{1}{p})\sin \alpha$$

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