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Question Number 42863 by ajfour last updated on 03/Sep/18

Commented by ajfour last updated on 04/Sep/18

$$Thetwoparabolashaveacommon$$$$vertexandhavethesamedistance$$$$fromvertextofocus{}^{\prime }{\mathit{a}}^{\prime }.$$$$Findtheangle\mathit{\theta }betweentheir$$$$axesiftheirbranchesthatare$$$$arrowmarkedalsomeetatinfinity.$$

Commented by MJS last updated on 04/Sep/18

$$\mathrm{they}\mathrm{will}\mathrm{always}\mathrm{intersect}\mathrm{in}\mathrm{a}\mathrm{real}\mathrm{point}\mathrm{except}$$$$\theta =180°$$

Commented by MrW3 last updated on 04/Sep/18

$$whatdoesitmean‘‘theymeetatinfinity$$$${also}^{″}?$$

Commented by ajfour last updated on 04/Sep/18

$$theyalreadyintersectatthevertex.$$$$Sothattheymeetalsoatx\to \mathrm{\infty },$$$$find\theta .$$

Commented by MJS last updated on 04/Sep/18

$$\mathrm{I}\mathrm{guess}\mathrm{the}\mathrm{idea}\mathrm{is},\mathrm{the}\mathrm{left}\mathrm{branch}\mathrm{of}\mathrm{the}\mathrm{red}$$$$\mathrm{one}\mathrm{meets}\mathrm{the}\mathrm{right}\mathrm{branch}\mathrm{of}\mathrm{the}\mathrm{blue}\mathrm{one}$$$$‘‘\mathrm{at}{\mathrm{infinity}}^{″},\mathrm{like}\mathrm{we}\mathrm{might}\mathrm{think}\mathrm{the}\mathrm{branches}$$$$\mathrm{of}\mathrm{a}\mathrm{hyperbola}\mathrm{cross}\mathrm{their}\mathrm{ways}‘‘\mathrm{at}{\mathrm{infinity}}^{″}$$

Commented by MJS last updated on 04/Sep/18

$$\mathrm{think}\mathrm{it}\mathrm{this}\mathrm{way}:$$$$\mathrm{start}\mathrm{with}\theta =90°\mathrm{and}\mathrm{then}\mathrm{slowly}\mathrm{turn}\mathrm{it}$$$$\mathrm{counterclockwise}\mathrm{towards}\theta =0°$$$${\mathrm{we}}^{\prime }\mathrm{ll}\mathrm{always}\mathrm{get}\mathrm{a}\mathrm{real}\mathrm{intersection}\mathrm{except}$$$$\mathrm{at}\theta =0°\mathrm{where}\mathrm{the}\mathrm{red}\mathrm{branch}‘‘\mathrm{loses}{\mathrm{contact}}^{″}$$$$\mathrm{of}\mathrm{the}\mathrm{right}\mathrm{blue}\mathrm{one}\mathrm{but}\mathrm{at}\mathrm{the}\mathrm{same}\mathrm{moment}$$$$\mathrm{becomes}\mathrm{one}\mathrm{with}\mathrm{the}\mathrm{left}\mathrm{one}...$$

Commented by MJS last updated on 04/Sep/18

$$\mathrm{to}\mathrm{see}\mathrm{what}\mathrm{happens}\mathrm{at}\mathrm{infinity}\mathrm{mirror}\mathrm{the}$$$$\mathrm{plain}\mathrm{at}\mathrm{the}\mathrm{circle}{x}^2+y^2=1$$$$P=(r,\theta )\to P^{\ast }=(\frac{1}{r},\theta )$$$$\left(\begin{array}{c}t\\ f\left(t\right)\end{array}\right)\to \left(\begin{array}{c}\frac{t}{t^2+f^2\left(t\right)}\\ \frac{f\left(t\right)}{t^2+f^2\left(t\right)}\end{array}\right)$$$$\mathrm{a}\mathrm{hyperbola}\mathrm{turns}\mathrm{into}\mathrm{a}\mathrm{lemniscate},\mathrm{it}\mathrm{has}$$$$\mathrm{got}2\mathrm{directions}(=\mathrm{the}\mathrm{directions}\mathrm{of}\mathrm{the}$$$$\mathrm{asymptotes})$$$$\mathrm{a}\mathrm{parabola}\mathrm{turns}\mathrm{into}\mathrm{a}\mathrm{round}\mathrm{shape}\mathrm{with}$$$$\mathrm{one}\mathrm{tip}\mathrm{in}\mathrm{the}\mathrm{center}\left(\begin{array}{c}0\\ 0\end{array}\right),\mathrm{it}\mathrm{has}\mathrm{got}\mathrm{only}\mathrm{one}$$$$\mathrm{direction},\mathrm{its}\mathrm{axis}\mathrm{of}\mathrm{symmetry}.\mathrm{you}\mathrm{can}\mathrm{imagine}$$$$\mathrm{it}\mathrm{as}\mathrm{an}\mathrm{ellipse}\mathrm{with}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{focus}\mathrm{in}\mathrm{infinity}$$

Answered by MJS last updated on 04/Sep/18

$$y={ax}^2$$$$x\to x\cos \theta -y\sin \theta$$$$y\to x\sin \theta +y\cos \theta$$$$y={ax}^2\to y=\frac{x}{\tan \theta }+\frac{\cos \theta }{a(1-\cos 2\theta )}\pm \frac{\sqrt{2(8ax\sin \theta +\cos 2\theta +1)}}{2a(1-\cos 2\theta )}$$$$\mathrm{and}\mathrm{this}\mathrm{is}\mathrm{defined}\mathrm{for}x⩾-\frac{1+\cos 2\theta }{8a\sin \theta }\mathrm{with}\theta >0$$$$\mathrm{so}\mathrm{no}\mathrm{matter}\mathrm{how}\mathrm{small}\mid \theta \mid \mathrm{is},\mathrm{the}\mathrm{left}\mathrm{branch}$$$$\mathrm{of}\mathrm{the}\mathrm{red}\mathrm{parabola}\mathrm{will}\mathrm{be}\mathrm{defined}\mathrm{in}$$$$[-\frac{1+\cos 2\theta }{8a\sin \theta };+\mathrm{\infty }[\Rightarrow \mathrm{it}\mathrm{will}\mathrm{intersect}\mathrm{the}\mathrm{right}$$$$\mathrm{branch}\mathrm{of}\mathrm{the}\mathrm{blue}\mathrm{parabola}$$

Answered by MrW3 last updated on 04/Sep/18

$${let}^{\prime }ssayeqn.oforiginalparabolais$$$$y={cx}^{2}withc=\frac{1}{4a}$$$$theeqn.ofrotatedparabolais$$$$x\sin \theta +y\cos \theta =c{(x\cos \theta -y\sin \theta )}^2$$$$$$$$intersectionpoint(h,k):$$$$h\sin \theta +{ch}^2\cos \theta =c{(h\cos \theta -{ch}^2\sin \theta )}^2$$$$\sin \theta +ch\cos \theta =ch{(\cos \theta -ch\sin \theta )}^2$$$$with\lambda =ch\ne 0$$$$\sin \theta +\lambda \cos \theta (1-\cos \theta )+2{\lambda }^2\sin \theta \cos \theta -{\lambda }^3{\sin }^2\theta =0$$$${\sin }^2\theta =\frac{\sin \theta }{{\lambda }^3}+\frac{\cos \theta (1-\cos \theta )}{{\lambda }^2}+\frac{\sin 2\theta }{\lambda }$$$$\lambda \to \mathrm{\infty }\Rightarrow \sin \theta \to 0$$$$i.e.suchthattheparabolasmeetatinfinity,$$$$\theta mustequal0.$$

Commented by ajfour last updated on 04/Sep/18

$$ThankyouSirMjS,SirMrW,$$$$iviewitbetterandcorrectnow$$$$withmy{mind}^{\prime }seyenow.All$$$$answersandcommentswere$$$$helpful.$$$$Aparabolagoesparalleltoits$$$$axismuchawayfromitsvertex.$$

Commented by MJS last updated on 04/Sep/18

$$\mathrm{as}\mathrm{always}{\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome}$$

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