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Question Number 127509 by mr W last updated on 30/Dec/20

Commented by mr W last updated on 31/Dec/20

$$ifbothellipseshavethesameshape$$$$as\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,determine\frac{b}{a}=?$$

Commented by mr W last updated on 28/Jan/21

Answered by mr W last updated on 31/Dec/20

Commented by mr W last updated on 25/Jan/21

$$letm=\tan \theta ,\mu =\frac{b}{a}$$$$\Rightarrow \sin \theta =\frac{m}{\sqrt{1+m^2}},\cos \theta =\frac{1}{\sqrt{1+m^2}}$$$$$$$$\mathit{i}\mathit{n}{\mathit{x}}^{\prime }{\mathit{y}}^{\prime }-\mathit{s}\mathit{y}\mathit{s}\mathit{t}\mathit{e}\mathit{m}:$$$$O(h,k)$$$$eqn.ofOA:$$$$y=k+m(x-h)$$$$\Rightarrow mx-y+k-mh=0$$$$OAistangenttoellipse:$$$$a^2{m}^2+b^2={(k-mh)}^2$$$$\Rightarrow k-mh=-\sqrt{a^2{m}^2+b^2}...\left(i\right)$$$$eqn.ofOB:$$$$y=k-\frac{1}{m}(x-h)$$$$\Rightarrow \frac{x}{m}+y-k-\frac{h}{m}=0$$$$OBistangenttoellipse:$$$$\frac{a^2}{m^2}+b^2={(k+\frac{h}{m})}^2$$$$\Rightarrow k+\frac{h}{m}=-\sqrt{\frac{a^2}{m^2}+b^2}...\left(ii\right)$$$$from\left(i\right)and\left(ii\right):$$$$\Rightarrow \frac{h}{a}=-\frac{1}{1+m^2}(\sqrt{1+m^2{\mu }^2}-m\sqrt{m^2+{\mu }^2})$$$$\Rightarrow \frac{k}{a}=-\frac{1}{1+m^2}(m\sqrt{1+m^2{\mu }^2}+\sqrt{m^2+{\mu }^2})$$$$sayP(a\cos \varphi ,-b\sin \varphi )$$$$\tan (\theta -\phi )=\frac{dy}{dx}=\frac{-b\cos \varphi }{-a\sin \varphi }=\frac{\mu }{\tan \varphi }$$$$\frac{m-\tan \phi }{1+m\tan \phi }=\frac{\mu }{\tan \varphi }$$$$let\eta =\tan \varphi$$$$\sin \varphi =\frac{\eta }{\sqrt{1+{\eta }^2}},\cos \varphi =\frac{1}{\sqrt{1+{\eta }^2}}$$$$\Rightarrow \tan \phi =\frac{\eta -\frac{\mu }{m}}{\frac{\eta }{m}+\mu }$$$$$$$$\mathit{i}\mathit{n}\mathit{x}\mathit{y}-\mathit{s}\mathit{y}\mathit{s}\mathit{t}\mathit{e}\mathit{m}:$$$$sayP(x_p,y_P)$$$$x_P=(-k-b\sin \varphi )\sin \theta +(-h+a\cos \varphi )\cos \theta$$$$\frac{x_P}{a}=[\frac{1}{1+m^2}(m\sqrt{1+m^2{\mu }^2}+\sqrt{m^2+{\mu }^2})-\mu \sin \varphi ]\sin \theta +[\frac{1}{1+m^2}(\sqrt{1+m^2{\mu }^2}-m\sqrt{m^2+{\mu }^2})+\cos \varphi ]\cos \theta$$$$\frac{x_P}{a}=\frac{m}{\sqrt{1+m^2}}[\frac{1}{1+m^2}(m\sqrt{1+m^2{\mu }^2}+\sqrt{m^2+{\mu }^2})-\frac{\mu \eta }{\sqrt{1+{\eta }^2}}]+\frac{1}{\sqrt{1+m^2}}[\frac{1}{1+m^2}(\sqrt{1+m^2{\mu }^2}-m\sqrt{m^2+{\mu }^2})+\frac{1}{\sqrt{1+{\eta }^2}}]$$$$\Rightarrow \frac{x_P}{a}=\frac{1}{\sqrt{1+m^2}}(\sqrt{1+m^2{\mu }^2}+\frac{1-m\mu \eta }{\sqrt{1+{\eta }^2}})$$$$$$$$y_P=(-k-b\sin \varphi )\cos \theta -(-h+a\cos \varphi )\sin \theta$$$$\frac{y_P}{a}=[\frac{1}{1+m^2}(m\sqrt{1+m^2{\mu }^2}+\sqrt{m^2+{\mu }^2})-\mu \sin \varphi ]\cos \theta -[\frac{1}{1+m^2}(\sqrt{1+m^2{\mu }^2}-m\sqrt{m^2+{\mu }^2})+\cos \varphi ]\sin \theta$$$$\frac{y_P}{a}=\frac{1}{\sqrt{1+m^2}}[\frac{1}{1+m^2}(m\sqrt{1+m^2{\mu }^2}+\sqrt{m^2+{\mu }^2})-\frac{\mu \eta }{\sqrt{1+{\eta }^2}}]-\frac{m}{\sqrt{1+m^2}}[\frac{1}{1+m^2}(\sqrt{1+m^2{\mu }^2}-m\sqrt{m^2+{\mu }^2})+\frac{1}{\sqrt{1+{\eta }^2}}]$$$$\Rightarrow \frac{y_P}{a}=\frac{1}{\sqrt{1+m^2}}(\sqrt{m^2+{\mu }^2}-\frac{m+\mu \eta }{\sqrt{1+{\eta }^2}})$$$$\frac{x_P^2}{a^2}+\frac{y_P^2}{b^2}=1$$$$\Rightarrow {(\sqrt{1+m^2{\mu }^2}+\frac{1-m\mu \eta }{\sqrt{1+{\eta }^2}})}^2+\frac{1}{{\mu }^2}{(\sqrt{m^2+{\mu }^2}-\frac{m+\mu \eta }{\sqrt{1+{\eta }^2}})}^2=1+m^2...\left(I\right)$$$$\tan \phi =-\frac{dy}{dx}={\left(\frac{b}{a}\right)}^2\frac{x_P}{y_P}$$$$\frac{\eta -\frac{\mu }{m}}{\frac{\eta }{m}+\mu }={\mu }^2\times \frac{\sqrt{1+m^2{\mu }^2}+\frac{1-m\mu \eta }{\sqrt{1+{\eta }^2}}}{\sqrt{m^2+{\mu }^2}-\frac{m+\mu \eta }{\sqrt{1+{\eta }^2}}}$$$$\Rightarrow {\mu }^2(\frac{\eta }{m}+\mu )(\sqrt{1+m^2{\mu }^2}+\frac{1-m\mu \eta }{\sqrt{1+{\eta }^2}})=(\eta -\frac{\mu }{m})(\sqrt{m^2+{\mu }^2}-\frac{m+\mu \eta }{\sqrt{1+{\eta }^2}})...\left(II\right)$$$$$$$$foranyvalueinrange0<\theta <60°we$$$$canfindthecoresponding\mu from$$$$\left(I\right)and\left(II\right).$$$$$$$$weget{\mu }_{max}\approx 0.38436919447at\theta \approx 37.9382°$$

Commented by mr W last updated on 31/Dec/20

Commented by mr W last updated on 31/Dec/20

Commented by mr W last updated on 31/Dec/20

Commented by mr W last updated on 02/Jan/21

Answered by mr W last updated on 30/Jan/21

Commented by mr W last updated on 30/Jan/21

$$letm=\tan \theta ,\mu =\frac{b}{a}$$$$\Rightarrow \sin \theta =\frac{m}{\sqrt{1+m^2}},\cos \theta =\frac{1}{\sqrt{1+m^2}}$$$$$$$$\mathit{i}\mathit{n}{\mathit{x}}^{\prime }{\mathit{y}}^{\prime }-\mathit{s}\mathit{y}\mathit{s}\mathit{t}\mathit{e}\mathit{m}:$$$$O(h,k)$$$$eqn.ofOA:$$$$y=k+m(x-h)$$$$\Rightarrow mx-y+k-mh=0$$$$OAistangenttoellipse:$$$$a^2{m}^2+b^2={(k-mh)}^2$$$$\Rightarrow k-mh=-\sqrt{a^2{m}^2+b^2}...\left(i\right)$$$$eqn.ofOB:$$$$y=k-\frac{1}{m}(x-h)$$$$\Rightarrow \frac{x}{m}+y-k-\frac{h}{m}=0$$$$OBistangenttoellipse:$$$$\frac{a^2}{m^2}+b^2={(k+\frac{h}{m})}^2$$$$\Rightarrow k+\frac{h}{m}=-\sqrt{\frac{a^2}{m^2}+b^2}...\left(ii\right)$$$$from\left(i\right)and\left(ii\right):$$$$\Rightarrow \frac{h}{a}=-\frac{1}{1+m^2}(\sqrt{1+m^2{\mu }^2}-m\sqrt{m^2+{\mu }^2})$$$$\Rightarrow \frac{k}{a}=-\frac{1}{1+m^2}(m\sqrt{1+m^2{\mu }^2}+\sqrt{m^2+{\mu }^2})$$$$$$$$\mathit{i}\mathit{n}\mathit{x}\mathit{y}-\mathit{s}\mathit{y}\mathit{s}\mathit{t}\mathit{e}\mathit{m}:$$$$say{O}^{\prime }(x_C,y_C)$$$$x_C=-k\sin \theta -h\cos \theta$$$$\frac{x_C}{a}=\frac{1}{1+m^2}(\sqrt{1+m^2{\mu }^2}-m\sqrt{m^2+{\mu }^2})\cos \theta +\frac{1}{1+m^2}(m\sqrt{1+m^2{\mu }^2}+\sqrt{m^2+{\mu }^2})\sin \theta$$$$\Rightarrow \frac{x_C}{a}=\frac{\sqrt{1+m^2{\mu }^2}}{\sqrt{1+m^2}}$$$$y_C=-k\cos \theta +h\sin \theta$$$$\frac{y_C}{a}=-\frac{1}{1+m^2}(\sqrt{1+m^2{\mu }^2}-m\sqrt{m^2+{\mu }^2})\sin \theta +\frac{1}{1+m^2}(m\sqrt{1+m^2{\mu }^2}+\sqrt{m^2+{\mu }^2})\cos \theta$$$$\Rightarrow \frac{y_C}{a}=\frac{\sqrt{m^2+{\mu }^2}}{\sqrt{1+m^2}}$$$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\Rightarrow {\mu }^2{\xi }^2+{\eta }^2={\mu }^2...\left(i\right)$$$$\Rightarrow \eta =\mu \sqrt{1-{\xi }^2}$$$$eqn.ofellipse{O}^{\prime }:$$$$\frac{{[(x-x_C)\frac{1}{\sqrt{1+m^2}}-(y-y_C)\frac{m}{\sqrt{1+m^2}}+x_C]}^2}{a^2}+\frac{{[(x-x_C)\sin \theta -(y-y_C)\cos \theta +y_C]}^2}{b^2}=1$$$${[(\frac{x}{a}-\frac{\sqrt{1+m^2{\mu }^2}}{\sqrt{1+m^2}})-m(\frac{y}{a}-\frac{\sqrt{m^2+{\mu }^2}}{\sqrt{1+m^2}})+\sqrt{1+m^2{\mu }^2}]}^2+\frac{{[m(\frac{x}{a}-\frac{\sqrt{1+m^2{\mu }^2}}{\sqrt{1+m^2}})+(\frac{y}{a}-\frac{\sqrt{m^2+{\mu }^2}}{\sqrt{1+m^2}})+\sqrt{m^2+{\mu }^2}]}^2}{{\mu }^2}=1+m^2$$$$\Rightarrow {\mu }^2{[\xi -m\eta +\frac{m\sqrt{m^2+{\mu }^2}-\sqrt{1+m^2{\mu }^2}}{\sqrt{1+m^2}}+\sqrt{1+m^2{\mu }^2}]}^2+{[m\xi +\eta -\frac{\sqrt{m^2+{\mu }^2}+m\sqrt{1+m^2{\mu }^2}}{\sqrt{1+m^2}}+\sqrt{m^2+{\mu }^2}]}^2={\mu }^2(1+m^2)...\left(ii\right)$$$$\Rightarrow {\mu }^2{(\xi -m\eta +p)}^2+{(m\xi +\eta +q)}^2={\mu }^2(1+m^2)...\left(ii\right)$$$$......$$

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