Find the equation of the hyperbola whose two foci are S(6,4) and S' (-4,4) and eccentricity is 2.
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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \mathrm{S}\left(6,4\right) \mathrm{and} \mathrm{S}\text{'}\left(-4,4\right) \mathrm{and} \mathrm{e}=2 \\ \mathrm{Slope} \mathrm{of} \mathrm{SS}\text{'}=\frac{4-4}{-4-6}=\frac{0}{-10}=0 \\ \mathrm{As} \mathrm{slope} \mathrm{of} \mathrm{hyperbola} \mathrm{is} 0, \mathrm{hence} \mathrm{axis} \mathrm{of} \mathrm{hyperbola} \mathrm{is} \mathrm{parallel} \mathrm{to} \mathrm{X}-\mathrm{axis}. \\ \mathrm{Equation} \mathrm{of} \mathrm{parabola} \mathrm{will} \mathrm{axis} \mathrm{parallel} \mathrm{to} \mathrm{X}-\mathrm{axis} \mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{{\left(\mathrm{x}-\mathrm{h}\right)}^2}{{\mathrm{a}}^2}-\frac{{\left(\mathrm{y}-\mathrm{k}\right)}^2}{{\mathrm{b}}^2}=1, \mathrm{where} \left(\mathrm{h},\mathrm{k}\right) \mathrm{is} \mathrm{the} \mathrm{center} \mathrm{of} \mathrm{hyperbola}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \mathrm{SS}\text{'}=2\mathrm{ae} \\ \Rightarrow \sqrt{{\left(-4-6\right)}^2+{\left(4-4\right)}^2}=4\mathrm{a} \\ \Rightarrow 4\mathrm{a}=\sqrt{100} \\ \Rightarrow 4\mathrm{a}=10 \\ \Rightarrow \mathrm{a}=\frac{5}{2} \\ \Rightarrow {\mathrm{a}}^2=\frac{25}{4} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ {\mathrm{b}}^2={\mathrm{a}}^2\left({\mathrm{e}}^2-1\right)=\left(\frac{25}{4}\right)\left(4-1\right)=\frac{75}{4} \\ \mathrm{Mid} \mathrm{point} \mathrm{of} \mathrm{S} \mathrm{and} \mathrm{S}\text{'} \mathrm{will} \mathrm{give} \mathrm{center} \mathrm{of} \mathrm{hyperbola}. \\ \Rightarrow \left(\mathrm{h},\mathrm{k}\right)=\left(\frac{6-4}{2},\frac{4+4}{2}\right)=\left(1,4\right) \\ \mathrm{Putting} \mathrm{all} \mathrm{values} \mathrm{in} \mathrm{equation} \frac{{\left(\mathrm{x}-\mathrm{h}\right)}^2}{{\mathrm{a}}^2}-\frac{{\left(\mathrm{y}-\mathrm{k}\right)}^2}{{\mathrm{b}}^2}=1, \mathrm{we} \mathrm{get}: \\ \frac{{\left(\mathrm{x}-1\right)}^2}{{\displaystyle \frac{25}{4}}}-\frac{{\left(\mathrm{y}-4\right)}^2}{{\displaystyle \frac{75}{4}}}=1 \\ \Rightarrow \frac{4{\left(\mathrm{x}-1\right)}^2}{{\displaystyle 25}}-\frac{4{\left(\mathrm{y}-4\right)}^2}{{\displaystyle 75}}=1 \\ \Rightarrow \frac{{\left(\mathbf{x}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}}{{\displaystyle 25}}\mathbf{-}\frac{{\left(\mathbf{y}\mathbf{-}\mathbf{4}\right)}^{\mathbf{2}}}{{\displaystyle 75}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}} \end{gathered}$$

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