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Question Number 66055 by ajfour last updated on 08/Aug/19

Commented by ajfour last updated on 08/Aug/19

$$Eq.ofellipse\frac{x^2}{a^2}+\frac{{(y-b)}^2}{b^2}=1$$$$parabolatouchesthex-axis$$$$andtheellipseandpasses$$$$thoughcenterofellipse(0,b).$$$$Findequationofparabolain$$$$termsofaandb.$$

Answered by mr W last updated on 09/Aug/19

$$eqn.ofparabola:$$$$y=b{(1-\frac{x}{p})}^2$$$$eqn.ofellipse:$$$${\left(\frac{x}{a}\right)}^2+{(\frac{y}{b}-1)}^2=1$$$$touchpointP(h,k)$$$$letm=\tan \theta fortangentattouchpoint$$$$y^{\prime }=-\frac{2b}{p}(1-\frac{h}{p})=m$$$$\Rightarrow h=p(1+\frac{pm}{2b})$$$$\frac{x}{a^2}+\frac{1}{b}(\frac{y}{b}-1)y^{\prime }=0$$$$\frac{h}{a^2}+\frac{1}{b}(\frac{k}{b}-1)m=0$$$$\Rightarrow k=b[1-\frac{pb}{{ma}^2}(1+\frac{pm}{2b})]$$$${\left(\frac{h}{a}\right)}^2+{(\frac{k}{b}-1)}^2=1$$$$\Rightarrow \frac{p^2}{a^2}(1+\frac{b^2}{m^2{a}^2}){(1+\frac{pm}{2b})}^2=1...\left(i\right)$$$$$$$$\frac{k}{b}={(1-\frac{h}{p})}^2$$$$\Rightarrow 1-\frac{pb}{{ma}^2}(1+\frac{pm}{2b})={\left(\frac{pm}{2b}\right)}^2...\left(ii\right)$$$$let\lambda =\frac{p}{a},\mu =\frac{b}{a}$$$$from\left(i\right):$$$$\Rightarrow {\lambda }^2(1+\frac{{\mu }^2}{m^2}){(1+\frac{m\lambda }{2\mu })}^2=1...\left(I\right)$$$$from\left(ii\right):$$$$\Rightarrow 1-\frac{\mu \lambda }{m}(1+\frac{m\lambda }{2\mu })={\left(\frac{m\lambda }{2\mu }\right)}^2$$$$\Rightarrow (2+\frac{m^2}{{\mu }^2}){\lambda }^2+\frac{4\mu }{m}\lambda -4=0...\left(II\right)$$$$\Rightarrow \lambda =\frac{\frac{2m}{\mu }}{2+\frac{m^2}{{\mu }^2}}=\frac{2\mu m}{2{\mu }^2+m^2}$$$$or{m}^2-\frac{2\mu }{\lambda }m+2{\mu }^2=0$$$$\Rightarrow \frac{m}{\mu }=\frac{1}{\lambda }(1+\sqrt{1-2{\lambda }^2})$$$$putthisinto\left(I\right):$$$$\Rightarrow \frac{{\lambda }^2{(3+\sqrt{1-2{\lambda }^2})}^2}{8}(2+\frac{{\lambda }^2}{1-{\lambda }^2+\sqrt{1-2{\lambda }^2}})=1$$$$\Rightarrow \lambda =0.5194(constant,independentfrom\mu !)$$$$\Rightarrow p=\lambda a=0.5194a$$$$$$$$eqn.ofparabola:$$$$\Rightarrow y=b{(1-\frac{x}{0.5194a})}^2$$

Commented by MJS last updated on 09/Aug/19

$$\mathrm{great}!$$$$\mathrm{just}\mathrm{for}\mathrm{the}\mathrm{fun}\mathrm{of}\mathrm{it},\mathrm{the}\mathrm{exact}\mathrm{value}\mathrm{of}\lambda :$$$$\lambda =\sqrt{-4+\sqrt[3]{56+\frac{88}{9}\sqrt{33}}-\sqrt[3]{-56+\frac{88}{9}\sqrt{33}}}$$

Commented by mr W last updated on 08/Aug/19

Commented by mr W last updated on 08/Aug/19

Commented by mr W last updated on 09/Aug/19

$$thankssir!howdidyougetthat?$$$$couldyoutransformtheeqn.intoa$$$$cubiceqn.?$$

Commented by MJS last updated on 12/Aug/19

$$\mathrm{sorry}\mathrm{I}\mathrm{get}\mathrm{no}\mathrm{notifications}\mathrm{again}...$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{transform}\mathrm{to}$$$${\lambda }^6+\mathcal{A}{\lambda }^4+\mathcal{B}{\lambda }^2+\mathcal{C}=0$$$$\mathrm{and}\mathrm{then}\mathrm{follow}\mathrm{the}\mathrm{usual}\mathrm{path}$$

Commented by mr W last updated on 14/Aug/19

$$nowiseesir!thanks!$$$$igetnonotificationeither.:($$

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