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Question Number 210457 by peter frank last updated on 09/Aug/24

Commented by peter frank last updated on 09/Aug/24

$$\left(\mathrm{b}\right)\mathrm{and}\left(\mathrm{c}\right)\mathrm{help}$$

Commented by mr W last updated on 10/Aug/24

$$\left(b\right)iswrong.$$$$thelocusis$$$$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2+b^2}$$

Commented by peter frank last updated on 11/Aug/24

$$\mathrm{thanks}\mathrm{sir}.\mathrm{please}\mathrm{help}$$

Answered by mr W last updated on 10/Aug/24

Commented by mr W last updated on 10/Aug/24

$$\left(c\right)$$$$saytheparametersoftheellipseare$$$$a,bwitha⩾band\mu =\frac{b}{a}⩽1.$$$$sayP(a\cos \theta ,b\sin \theta )$$$$\tan \alpha =\frac{b\tan \theta }{a}=\mu \tan \theta$$$$\tan \beta =\frac{b}{a\tan \theta }=\frac{\mu }{\tan \theta }$$$$sayQ(a\cos \phi ,-b\sin \phi )$$$$\tan \beta =\frac{b\tan \phi }{a}=\mu \tan \phi$$$$\tan \alpha =\frac{b}{a\tan \phi }=\frac{\mu }{\tan \phi }$$$$\Rightarrow \tan \theta \tan \phi =1$$$$\mathrm{\Phi }=\tan (\alpha +\beta )$$$$=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$$$$=\frac{\mu \tan \theta +\mu \tan \phi }{1-{\mu }^2\tan \theta \tan \phi }$$$$=\frac{\mu (\tan \theta +\frac{1}{\tan \theta })}{1-{\mu }^2}⩾\frac{2\mu }{1-{\mu }^2}$$$${\mathrm{\Phi }}_{min}=\frac{2\mu }{1-{\mu }^2}when\tan \theta =\frac{1}{\tan \theta },i.e.$$$$\tan \theta =1=\tan \phi ,or\theta =\phi =\frac{\pi }{4}$$$$\theta =\phi meansthattheconjugate$$$$diametersareequal.$$$${(\alpha +\beta )}_{min}={\tan }^{-1}\frac{2\mu }{1-{\mu }^2}$$

Commented by hardmath last updated on 11/Aug/24

$$$$Dear professor, your solutions are perfect

Commented by mr W last updated on 11/Aug/24

$$thanks!$$

Commented by peter frank last updated on 14/Aug/24

$$\mathrm{thank}\mathrm{you}$$

Answered by mr W last updated on 11/Aug/24

$$\left(c\right)$$$$saythepoleisP(u,v).$$$$thepolarofPwithrespecttothe$$$$hyperbola\frac{x^2}{a^2}-\frac{y^2}{b^2}=1is$$$$\frac{ux}{a^2}-\frac{vy}{b^2}=1$$$$suchthatittangentsthecircle$$$$x^2+y^2=r^2(withr=ae=\sqrt{a^2+b^2}),$$$$thedistancefrom(0,0)to$$$$\frac{ux}{a^2}-\frac{vy}{b^2}=1mustber,i.e.$$$$\frac{1}{\sqrt{{\left(\frac{u}{a^2}\right)}^2+{(-\frac{v}{b^2})}^2}}=r=\sqrt{a^2+b^2}$$$$\Rightarrow \frac{u^2}{a^4}+\frac{v^2}{b^4}=\frac{1}{a^2+b^2}$$$$i.e.thelocusofthepoleis$$$$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2+b^2}$$

Commented by peter frank last updated on 11/Aug/24

$$\mathrm{thank}\mathrm{you}$$

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