find the equation of hyperbola in standard form having  distance between directrices is  4/3 root3  and passing through point (2,1)????????

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$$\begin{gathered} \mathrm{Standard} \mathrm{equation} \mathrm{of} \mathrm{hyperbola} \mathrm{is}: \\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{As} \mathrm{it} \mathrm{passes} \mathrm{through} \left(2,1\right). \mathrm{Hence} \\ \frac{4}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}=1 \\ \Rightarrow \frac{4}{{\mathrm{a}}^2}-1=\frac{1}{{\mathrm{b}}^2} \\ \Rightarrow \frac{4-{\mathrm{a}}^2}{{\mathrm{a}}^2}=\frac{1}{{\mathrm{b}}^2} \\ \Rightarrow {\mathrm{b}}^2=\frac{{\mathrm{a}}^2}{4-{\mathrm{a}}^2} ;\left(\mathrm{i}\right) \\ \mathrm{Distance} \mathrm{between} \mathrm{directrices}=\frac{4\sqrt{3}}{3} \\ \Rightarrow 2\mathrm{ae}=\frac{4\sqrt{3}}{3} \\ \Rightarrow 4{\mathrm{a}}^2{\mathrm{e}}^2=\frac{16\times 3}{9} \\ \Rightarrow {\mathrm{a}}^2{\mathrm{e}}^2=\frac{4}{3} \left(\mathrm{ii}\right) \\ \mathrm{Also} \\ {\mathrm{e}}^2=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2} \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{a}}^2{\mathrm{e}}^2 \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=\frac{4}{3} \left[\mathrm{Using} \left(\mathrm{ii}\right)\right] \\ {\mathrm{a}}^2+\frac{{\mathrm{a}}^2}{4-{\mathrm{a}}^2}=\frac{4}{3} \left[\mathrm{Using} \left(\mathrm{i}\right)\right] \\ \Rightarrow \frac{4{\mathrm{a}}^2-{\mathrm{a}}^4+{\mathrm{a}}^2}{4-{\mathrm{a}}^2}=\frac{4}{3} \\ \Rightarrow \frac{5{\mathrm{a}}^2-{\mathrm{a}}^4}{4-{\mathrm{a}}^2}=\frac{4}{3} \\ \Rightarrow 15{\mathrm{a}}^2-3{\mathrm{a}}^4=16-4{\mathrm{a}}^2 \\ \Rightarrow 3{\mathrm{a}}^4-19{\mathrm{a}}^2+16=0 \\ \Rightarrow 3{\mathrm{a}}^4-3{\mathrm{a}}^2-16{\mathrm{a}}^2+16=0 \\ \Rightarrow 3{\mathrm{a}}^2\left({\mathrm{a}}^2-1\right)-16\left({\mathrm{a}}^2-1\right)=0 \\ \Rightarrow \left({\mathrm{a}}^2-1\right)\left(3{\mathrm{a}}^2-1\right)=0 \\ \Rightarrow {\mathrm{a}}^2=1 \mathrm{or} {\mathrm{a}}^2=\frac{1}{3} \\ \mathrm{When} {\mathrm{a}}^2=1 \\ {\mathrm{b}}^2=\frac{1}{4-1}=\frac{1}{3} \\ \mathrm{Hence} \mathrm{equation} \mathrm{will} \mathrm{be}: \\ \frac{{\mathrm{x}}^2}{1}-\frac{{\mathrm{y}}^2}{\frac{1}{3}}=1 \\ \Rightarrow \boxed{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{1}} \\ \mathrm{When} {\mathrm{a}}^2=\frac{1}{3} \\ {\mathrm{b}}^2=\frac{{\displaystyle \frac{1}{3}}}{4-{\displaystyle \frac{1}{3}}} \\ \Rightarrow {\mathrm{b}}^2=\frac{1}{11} \\ \mathrm{Hence} \mathrm{equation} \mathrm{will} \mathrm{be}: \\ \frac{{\mathrm{x}}^2}{{\displaystyle \frac{1}{3}}}-\frac{{\mathrm{y}}^2}{{\displaystyle \frac{1}{11}}}=1 \\ \Rightarrow \boxed{\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{11}{\mathbf{y}}^{\mathbf{2}}\mathbf{=}\mathbf{1}} \end{gathered}$$

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