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Question Number 127460 by ajfour last updated on 30/Dec/20

Commented by ajfour last updated on 30/Dec/20

$$Q.127127\left(Revisit\right)$$$$Givenp,q;findR.$$

Answered by ajfour last updated on 30/Dec/20

$$\frac{{(x-p)}^2}{p^2}+\frac{{(y-q)}^2}{q^2}=1$$$$slopeoftangent$$$$-m=-\frac{\cos \theta }{\sin \theta }=-\frac{q^2}{p^2}\left(\frac{x-p}{y-q}\right)$$$$VariablepointQ(R\cos \theta ,R\sin \theta )$$$$m=\frac{\cos \theta }{\sin \theta }=\frac{q^2(p-R\cos \theta )}{p^2(q-R\sin \theta )}$$$$\Rightarrow p^{2}q\cos \theta -p^{2}R\sin \theta \cos \theta$$$$={pq}^2\sin \theta -q^{2}R\sin \theta \cos \theta$$$$\Rightarrow pq(p\cos \theta -q\sin \theta )$$$$=(p^2-q^2)R\sin \theta \cos \theta$$$$R=\frac{pq}{(p^2-q^2)}(\frac{p}{\sin \theta }-\frac{q}{\cos \theta })$$$$s^2={(p-p\cos \varphi -R)}^2+{(q-q\sin \varphi -R)}^2$$$$\frac{{ds}^2}{d\varphi }=0\Rightarrow$$$$p(p-p\cos \varphi -R)\sin \varphi =q(q-q\sin \varphi -R)\cos \varphi$$$$\Rightarrow \frac{q-q\sin \varphi -R}{p-p\cos \varphi -R}=\frac{p\sin \varphi }{q\cos \varphi }$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$R(p\sin \varphi -q\cos \varphi )=(p^2-q^2)\sin \varphi \cos \varphi$$$$-(p^2\sin \varphi -q^2\cos \varphi )$$$$R^2={(p-p\cos \varphi -R)}^2+{(q-q\sin \varphi -R)}^2$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$$$

Answered by mr W last updated on 30/Dec/20

$$Q(p-p\cos \phi ,q-q\sin \phi )$$$$p^2{(1-\cos \phi )}^2+q^2{(1-\sin \phi )}^2=R^2$$$$\frac{p(1-\cos \phi )}{q(1-\sin \phi )}=\frac{q\cos \phi }{p\sin \phi }$$$$let\mu =\frac{q}{p},\lambda =\frac{R}{p}$$$$\frac{\sin \phi (1-\cos \phi )}{\cos \phi (1-\sin \phi )}={\mu }^2...\left(i\right)$$$${\lambda }^2={(1-\cos \phi )}^2+{\mu }^2{(1-\sin \phi )}^2...\left(ii\right)$$

Commented by mr W last updated on 30/Dec/20

Commented by mr W last updated on 30/Dec/20

$$ifailedtoeliminate\phi fromthe$$$$equations.$$

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