Question Number 84689 by mr W last updated on 15/Mar/20
Commented by mr W last updated on 15/Mar/20
$${the}\:{parabola}\:{is}\:{rolled}\:{along}\:{the}\:{circle} \\ $$$${as}\:{shown}. \\ $$$${find}\:{the}\:{equation}\:{of}\:{the}\:{parabola}\:{when} \\ $$$${the}\:{touching}\:{point}\:{is}\:{at}\:{the}\:{position}\:\theta. \\ $$
Commented by mr W last updated on 16/Mar/20
Answered by mr W last updated on 15/Mar/20
Commented by mr W last updated on 16/Mar/20
$${say}\:{in}\:{local}\:{system}\:{x}'{y}'\:{the}\:{point}\:{P} \\ $$$${is}\:\left({p},\:{p}^{\mathrm{2}} \right). \\ $$$${BP}={AP}={a}\theta \\ $$$${BP}=\int_{\mathrm{0}} ^{\:{p}} \sqrt{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} }\:{dx}=\frac{\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{2}{p}\right)+\left(\mathrm{2}{p}\right)\sqrt{\mathrm{1}+\left(\mathrm{2}{p}\right)^{\mathrm{2}} }}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{2}{p}\right)+\left(\mathrm{2}{p}\right)\sqrt{\mathrm{1}+\left(\mathrm{2}{p}\right)^{\mathrm{2}} }=\mathrm{4}{a}\theta \\ $$$$\mathrm{tan}\:\varphi={y}'=\mathrm{2}{p} \\ $$$$\Rightarrow\varphi=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{p}\right) \\ $$$$ \\ $$$${in}\:{xy}−{system}: \\ $$$${P}\left({a}\:\mathrm{sin}\:\theta,\:{a}\:\mathrm{cos}\:\theta\right) \\ $$$${inclination}\:{angle}\:{of}\:{x}'−{axis}: \\ $$$$\alpha=\pi−\theta−\varphi \\ $$$${x}_{{B}} ={a}\:\mathrm{sin}\:\theta−{p}^{\mathrm{2}} \:\mathrm{sin}\:\alpha+{p}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow{x}_{{B}} ={a}\:\mathrm{sin}\:\theta−{p}^{\mathrm{2}} \:\mathrm{sin}\:\left(\theta+\varphi\right)−{p}\:\mathrm{cos}\:\left(\theta+\varphi\right) \\ $$$${y}_{{B}} ={a}\:\mathrm{cos}\:\theta+{p}^{\mathrm{2}} \:\mathrm{cos}\:\alpha+{p}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow{y}_{{B}} ={a}\:\mathrm{cos}\:\theta−{p}^{\mathrm{2}} \:\mathrm{cos}\:\left(\theta+\varphi\right)+{p}\:\mathrm{sin}\:\left(\theta+\varphi\right) \\ $$$${x}_{{F}} ={x}_{{B}} +\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow{x}_{{F}} ={a}\:\mathrm{sin}\:\theta+\left(\frac{\mathrm{1}}{\mathrm{4}}−{p}^{\mathrm{2}} \right)\:\mathrm{sin}\:\left(\theta+\varphi\right)−{p}\:\mathrm{cos}\:\left(\theta+\varphi\right) \\ $$$${y}_{{F}} ={y}_{{B}} −\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow{y}_{{F}} ={a}\:\mathrm{cos}\:\theta+\left(\frac{\mathrm{1}}{\mathrm{4}}−{p}^{\mathrm{2}} \right)\:\mathrm{cos}\:\left(\theta+\varphi\right)+{p}\:\mathrm{sin}\:\left(\theta+\varphi\right) \\ $$$${similarly} \\ $$$$\Rightarrow{x}_{{E}} ={a}\:\mathrm{sin}\:\theta−\left(\frac{\mathrm{1}}{\mathrm{4}}+{p}^{\mathrm{2}} \right)\:\mathrm{sin}\:\left(\theta+\varphi\right)−{p}\:\mathrm{cos}\:\left(\theta+\varphi\right) \\ $$$$\Rightarrow{y}_{{E}} ={a}\:\mathrm{cos}\:\theta−\left(\frac{\mathrm{1}}{\mathrm{4}}+{p}^{\mathrm{2}} \right)\:\mathrm{cos}\:\left(\theta+\varphi\right)+{p}\:\mathrm{sin}\:\left(\theta+\varphi\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{directix}: \\ $$$${y}−{y}_{{E}} =\mathrm{tan}\:\alpha\:\left({x}−{x}_{{E}} \right) \\ $$$$\mathrm{tan}\:\left(\theta+\varphi\right)\left({x}−{x}_{{E}} \right)+{y}−{y}_{{E}} =\mathrm{0} \\ $$$$ \\ $$$${eqn}.\:{of}\:{rolled}\:{parabola}\:{in}\:{xy}−{system}: \\ $$$${distance}\:{to}\:{directrix}\:=\:{distance}\:{to}\:{focus} \\ $$$$\frac{\mathrm{tan}\:\left(\theta+\varphi\right)\left({x}−{x}_{{E}} \right)+{y}−{y}_{{E}} }{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \:\left(\theta+\varphi\right)+\mathrm{1}}}=\sqrt{\left({x}−{x}_{{F}} \right)^{\mathrm{2}} +\left({y}−{y}_{{F}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left[\mathrm{tan}\:\left(\theta+\varphi\right)\left({x}−{x}_{{E}} \right)+{y}−{y}_{{E}} \right]^{\mathrm{2}} =\left[\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\theta+\varphi\right)\right]\left[\left({x}−{x}_{{F}} \right)^{\mathrm{2}} +\left({y}−{y}_{{F}} \right)^{\mathrm{2}} \right] \\ $$
Commented by mr W last updated on 16/Mar/20
Commented by mr W last updated on 16/Mar/20
Commented by mr W last updated on 16/Mar/20
Commented by mr W last updated on 16/Mar/20
Commented by mr W last updated on 16/Mar/20
Commented by mr W last updated on 17/Mar/20
$${find}\:{the}\:{parabola}\:{with}\:{its}\:{initial} \\ $$$${orientation} \\ $$$$ \\ $$$$\alpha=\pi−\theta−\varphi=−\pi \\ $$$$\Rightarrow\theta=\mathrm{2}\pi−\varphi=\mathrm{2}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{p}\right) \\ $$$$\Rightarrow\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{2}{p}\right)+\left(\mathrm{2}{p}\right)\sqrt{\mathrm{1}+\left(\mathrm{2}{p}\right)^{\mathrm{2}} }=\mathrm{4}{a}\left[\mathrm{2}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{p}\right)\right] \\ $$
Commented by mr W last updated on 17/Mar/20