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 (p, p^2 ). BP=AP=aθ BP=∫_0 ^( p) (√(1+(2x)^2 )) dx=((sinh^(−1) (2p)+(2p)(√(1+(2p)^2 )))/4) ⇒sinh^(−1) (2p)+(2p)(√(1+(2p)^2 ))=4aθ tan ϕ=y′=2p ⇒ϕ=tan^(−1) (2p) in xy−system: P(a sin θ, a cos θ) inclination angle of x′−axis: α=π−θ−ϕ x_B =a sin θ−p^2 sin α+p cos α ⇒x_B =a sin θ−p^2 sin (θ+ϕ)−p cos (θ+ϕ) y_B =a cos θ+p^2 cos α+p sin α ⇒y_B =a cos θ−p^2 cos (θ+ϕ)+p sin (θ+ϕ) x_F =x_B +(1/4) sin α ⇒x_F =a sin θ+((1/4)−p^2 ) sin (θ+ϕ)−p cos (θ+ϕ) y_F =y_B −(1/4) cos α ⇒y_F =a cos θ+((1/4)−p^2 ) cos (θ+ϕ)+p sin (θ+ϕ) similarly ⇒x_E =a sin θ−((1/4)+p^2 ) sin (θ+ϕ)−p cos (θ+ϕ) ⇒y_E =a cos θ−((1/4)+p^2 ) cos (θ+ϕ)+p sin (θ+ϕ) eqn. of directix: y−y_E =tan α (x−x_E ) tan (θ+ϕ)(x−x_E )+y−y_E =0 eqn. of rolled parabola in xy−system: distance to directrix = distance to focus ((tan (θ+ϕ)(x−x_E )+y−y_E )/( (√(tan^2 (θ+ϕ)+1))))=(√((x−x_F )^2 +(y−y_F )^2 )) ⇒[tan (θ+ϕ)(x−x_E )+y−y_E ]^2 =[1+tan^2 (θ+ϕ)][(x−x_F )^2 +(y−y_F )^2 ]](https://www.tinkutara.com/question/Q84829.png)
$${say}\:{in}\:{local}\:{system}\:{x}'{y}'\:{the}\:{point}\:{P} \\ $$$${is}\:\left({p},\:{p}^{\mathrm{2}} \right). \\ $$$${BP}={AP}={a}\theta \\ $$$${BP}=\int_{\mathrm{0}} ^{\:{p}} \sqrt{\mathrm{1}+\left(\mathrm{2}{x}\right)^{\mathrm{2}} }\:{dx}=\frac{\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{2}{p}\right)+\left(\mathrm{2}{p}\right)\sqrt{\mathrm{1}+\left(\mathrm{2}{p}\right)^{\mathrm{2}} }}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{2}{p}\right)+\left(\mathrm{2}{p}\right)\sqrt{\mathrm{1}+\left(\mathrm{2}{p}\right)^{\mathrm{2}} }=\mathrm{4}{a}\theta \\ $$$$\mathrm{tan}\:\varphi={y}'=\mathrm{2}{p} \\ $$$$\Rightarrow\varphi=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{p}\right) \\ $$$$ \\ $$$${in}\:{xy}−{system}: \\ $$$${P}\left({a}\:\mathrm{sin}\:\theta,\:{a}\:\mathrm{cos}\:\theta\right) \\ $$$${inclination}\:{angle}\:{of}\:{x}'−{axis}: \\ $$$$\alpha=\pi−\theta−\varphi \\ $$$${x}_{{B}} ={a}\:\mathrm{sin}\:\theta−{p}^{\mathrm{2}} \:\mathrm{sin}\:\alpha+{p}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow{x}_{{B}} ={a}\:\mathrm{sin}\:\theta−{p}^{\mathrm{2}} \:\mathrm{sin}\:\left(\theta+\varphi\right)−{p}\:\mathrm{cos}\:\left(\theta+\varphi\right) \\ $$$${y}_{{B}} ={a}\:\mathrm{cos}\:\theta+{p}^{\mathrm{2}} \:\mathrm{cos}\:\alpha+{p}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow{y}_{{B}} ={a}\:\mathrm{cos}\:\theta−{p}^{\mathrm{2}} \:\mathrm{cos}\:\left(\theta+\varphi\right)+{p}\:\mathrm{sin}\:\left(\theta+\varphi\right) \\ $$$${x}_{{F}} ={x}_{{B}} +\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow{x}_{{F}} ={a}\:\mathrm{sin}\:\theta+\left(\frac{\mathrm{1}}{\mathrm{4}}−{p}^{\mathrm{2}} \right)\:\mathrm{sin}\:\left(\theta+\varphi\right)−{p}\:\mathrm{cos}\:\left(\theta+\varphi\right) \\ $$$${y}_{{F}} ={y}_{{B}} −\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{cos}\:\alpha \\ $$$$\Rightarrow{y}_{{F}} ={a}\:\mathrm{cos}\:\theta+\left(\frac{\mathrm{1}}{\mathrm{4}}−{p}^{\mathrm{2}} \right)\:\mathrm{cos}\:\left(\theta+\varphi\right)+{p}\:\mathrm{sin}\:\left(\theta+\varphi\right) \\ $$$${similarly} \\ $$$$\Rightarrow{x}_{{E}} ={a}\:\mathrm{sin}\:\theta−\left(\frac{\mathrm{1}}{\mathrm{4}}+{p}^{\mathrm{2}} \right)\:\mathrm{sin}\:\left(\theta+\varphi\right)−{p}\:\mathrm{cos}\:\left(\theta+\varphi\right) \\ $$$$\Rightarrow{y}_{{E}} ={a}\:\mathrm{cos}\:\theta−\left(\frac{\mathrm{1}}{\mathrm{4}}+{p}^{\mathrm{2}} \right)\:\mathrm{cos}\:\left(\theta+\varphi\right)+{p}\:\mathrm{sin}\:\left(\theta+\varphi\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{directix}: \\ $$$${y}−{y}_{{E}} =\mathrm{tan}\:\alpha\:\left({x}−{x}_{{E}} \right) \\ $$$$\mathrm{tan}\:\left(\theta+\varphi\right)\left({x}−{x}_{{E}} \right)+{y}−{y}_{{E}} =\mathrm{0} \\ $$$$ \\ $$$${eqn}.\:{of}\:{rolled}\:{parabola}\:{in}\:{xy}−{system}: \\ $$$${distance}\:{to}\:{directrix}\:=\:{distance}\:{to}\:{focus} \\ $$$$\frac{\mathrm{tan}\:\left(\theta+\varphi\right)\left({x}−{x}_{{E}} \right)+{y}−{y}_{{E}} }{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \:\left(\theta+\varphi\right)+\mathrm{1}}}=\sqrt{\left({x}−{x}_{{F}} \right)^{\mathrm{2}} +\left({y}−{y}_{{F}} \right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left[\mathrm{tan}\:\left(\theta+\varphi\right)\left({x}−{x}_{{E}} \right)+{y}−{y}_{{E}} \right]^{\mathrm{2}} =\left[\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\theta+\varphi\right)\right]\left[\left({x}−{x}_{{F}} \right)^{\mathrm{2}} +\left({y}−{y}_{{F}} \right)^{\mathrm{2}} \right] \\ $$
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 α=π−θ−ϕ=−π ⇒θ=2π−ϕ=2π−tan^(−1) (2p) ⇒sinh^(−1) (2p)+(2p)(√(1+(2p)^2 ))=4a[2π−tan^(−1) (2p)]](https://www.tinkutara.com/question/Q84895.png)
$${find}\:{the}\:{parabola}\:{with}\:{its}\:{initial} \\ $$$${orientation} \\ $$$$ \\ $$$$\alpha=\pi−\theta−\varphi=−\pi \\ $$$$\Rightarrow\theta=\mathrm{2}\pi−\varphi=\mathrm{2}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{p}\right) \\ $$$$\Rightarrow\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{2}{p}\right)+\left(\mathrm{2}{p}\right)\sqrt{\mathrm{1}+\left(\mathrm{2}{p}\right)^{\mathrm{2}} }=\mathrm{4}{a}\left[\mathrm{2}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}{p}\right)\right] \\ $$
Commented by mr W last updated on 17/Mar/20
