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Question Number 39860 by ajfour last updated on 12/Jul/18

Commented by ajfour last updated on 12/Jul/18

$$OA=BP,OT=1,\mathrm{\&}$$$$\mathrm{\angle }OTB=\mathrm{\angle }BTP\left(givenconditions\right).$$

Answered by ajfour last updated on 13/Jul/18

$$\frac{{xx}_T}{a^2}+\frac{{yy}_T}{b^2}=1(eq.oftangentatT)$$$$(0,b+a)liesonthistangent$$$$\Rightarrow {\mathit{y}}_{\mathit{T}}=\frac{{\mathit{b}}^2}{\mathit{a}+\mathit{b}}.....\left(\mathit{i}\right)$$$$Andsince\frac{x_T^2}{a^2}+\frac{y_T^2}{b^2}=1,hence$$$${\mathit{x}}_{\mathit{T}}^2={\mathit{a}}^2-\frac{{\mathit{a}}^2{\mathit{b}}^2}{{(\mathit{a}+\mathit{b})}^2}...\left(\mathit{i}\mathit{i}\right)$$$$AsOT=1so{\mathit{x}}_{\mathit{T}}^2+{\mathit{y}}_{\mathit{T}}^2=1$$$$\Rightarrow a^2-\frac{a^2{b}^2}{{(a+b)}^2}+\frac{b^4}{{(a+b)}^2}=1...\left(iii\right)$$$$(a^2-1)=\frac{b^2(a^2-b^2)}{{(a+b)}^2}$$$$onesolutionofthiseq.is$$$$a=b=1,otherwise$$$$a^2-1=\frac{b^2(a-b)}{a+b}$$$$ifa=brthen$$$$b^2{r}^2-1=b^2\left(\frac{r-1}{r+1}\right)$$$$b^2(r^2-\frac{r-1}{r+1})=1...\left(iv\right)$$$$Andas{(\frac{PT}{OT}=\frac{BP}{OB})}^2$$$$\Rightarrow \frac{x_T^2+{[(a+b)-y_T]}^2}{1}=\frac{a^2}{b^2}$$$$\Rightarrow a^2-\frac{a^2{b}^2}{{(a+b)}^2}+{(a+b)}^2+y_T^2$$$$-2(a+b)y_T=\frac{a^2}{b^2}$$$$\Rightarrow a^2-\frac{a^2{b}^2}{{(a+b)}^2}+{(a+b)}^2+\frac{b^4}{{(a+b)}^2}$$$$-2b^2=\frac{a^2}{b^2}...\left(v\right)$$$$using\left(iii\right)inaboveeq.\left(v\right)$$$$1+{(\mathit{a}+\mathit{b})}^2-2{\mathit{b}}^2=\frac{{\mathit{a}}^2}{{\mathit{b}}^2}$$$$againwitha=br$$$$1+b^2{(r+1)}^2-2b^2=r^2$$$$\Rightarrow b^2(r^2+2r-1)=r^2-1...\left(vi\right)$$$$comparing\left(iv\right)and\left(vi\right)$$$$\Rightarrow \frac{r^2+2r-1}{r^2-\frac{r-1}{r+1}}=\frac{r^2-1}{1}$$$$\Rightarrow \frac{2r}{r^2-1-\frac{r-1}{r+1}}=r^2-1$$$$2r(r+1)=(r^2-1)[(r^2-1)(r+1)-(r-1)]$$$$\Rightarrow 2r={(r-1)}^2[{(r+1)}^2-1]$$$$\Rightarrow 2r={(r^2-1)}^2-{(r-1)}^2$$$$or{\mathit{r}}^4-3{\mathit{r}}^2=0$$$$\Rightarrow r=\sqrt{3}\left(validanswer\right)$$$$Nowfrom\left(vi\right)wehave$$$${\mathit{b}}^2=\frac{{\mathit{r}}^2-1}{{\mathit{r}}^2+2\mathit{r}-1}=\frac{2}{2+2\sqrt{3}}=\frac{1}{1+\sqrt{3}}$$$$=\frac{\sqrt{3}-1}{2}$$$$While{\mathit{a}}^2={\mathit{b}}^2{\mathit{r}}^2=\frac{3(\sqrt{3}-1)}{2}$$$$Ellipseeq.isthus$$$$\frac{2x^2}{3(\sqrt{3}-1)}+\frac{2y^2}{\sqrt{3}-1}=1$$$$or\underset{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}{}$$$$2x^2+6y^2=3\sqrt{3}-3$$$$\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.}{}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

$$bah...verygood...excellent...$$

Commented by ajfour last updated on 13/Jul/18

$$thankssir!$$

Commented by MrW3 last updated on 13/Jul/18

$$awesome!$$

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