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Question Number 48246 by ajfour last updated on 21/Nov/18

Commented by ajfour last updated on 21/Nov/18

$$Finda/b,iftheellipseparameters$$$$areevenaandb.$$

Answered by mr W last updated on 21/Nov/18

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\frac{x}{a^2}+\frac{{yy}^{\prime }}{b^2}=0$$$$P(h,k)$$$$y^{\prime }=-\frac{2b}{2a}=-\frac{b}{a}$$$$\frac{h}{2b}=\frac{2a}{2\sqrt{a^2+b^2}}$$$$\Rightarrow h=\frac{2ab}{\sqrt{a^2+b^2}}$$$$\frac{h}{a^2}-\frac{k}{ab}=0$$$$\Rightarrow k=\frac{bh}{a}=\frac{2b^2}{\sqrt{a^2+b^2}}$$$$\frac{h^2}{a^2}+\frac{k^2}{b^2}=1$$$$\frac{4a^2{b}^2}{a^2(a^2+b^2)}+\frac{4b^4}{b^2(a^2+b^2)}=1$$$$\frac{8b^2}{a^2+b^2}=1$$$$7b^2=a^2$$$$\Rightarrow \frac{a}{b}=\sqrt{7}=2.65$$

Commented by ajfour last updated on 21/Nov/18

$$MuchbettersolutionSir!$$

Answered by ajfour last updated on 21/Nov/18

$$LetP(h,k)$$$$eq.oftangent$$$$\frac{hx}{a^2}+\frac{ky}{b^2}=1$$$$slope=-\tan \theta =-\frac{b^2}{a^2}\left(\frac{h}{k}\right)..\left(i\right)$$$$Ifangleoftangentwithx-axes$$$$is\pi -\theta ,then$$$$h=2b\cos \theta =2a\sin \theta$$$$h^2(\frac{1}{a^2}+\frac{1}{b^2})=4....\left(ii\right)$$$$\Rightarrow \tan \theta =\frac{b}{a}...\left(iii\right)$$$$from\left(i\right)\mathrm{\&}\left(iii\right)$$$$\frac{b^2}{a^2}\left(\frac{h}{k}\right)=\frac{b}{a}\Rightarrow \frac{h}{k}=\frac{a}{b}...\left(iv\right)$$$$Also\frac{h^2}{a^2}+\frac{k^2}{b^2}=1$$$$\Rightarrow h^2(\frac{1}{a^2}+\frac{k^2}{h^2}\times \frac{1}{b^2})=1$$$$using\left(iii\right)$$$$h^2(\frac{1}{a^2}+\frac{1}{a^2})=1$$$$\Rightarrow 2h^2=a^2......\left(v\right)$$$$using\left(v\right)in\left(ii\right)$$$$\frac{a^2}{2}(\frac{1}{a^2}+\frac{1}{b^2})=4$$$$\Rightarrow 1+\frac{a^2}{b^2}=8$$$$or\frac{a}{b}=\sqrt{\stackrel{}{7}}.$$

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