Question

if e₁ be the eccentricity of the ellipse x²/16+y²/25=1 and e₂ is the eccentricity of the hyperbola passing through the foci of the ellipse and e₁e₂=1 then equation of hyperbola is

Answer 1

$$\begin{gathered} \frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{25}=1 \\ \mathrm{e}=\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}}{\mathrm{a}} \\ {\mathrm{e}}_1=\frac{\sqrt{25-16}}{5}=\frac{3}{5} \\ {\mathrm{e}}_1{\mathrm{e}}_2=1 \\ \frac{3}{5}\times {\mathrm{e}}_2=1 \\ {\mathrm{e}}_2=\frac{5}{3} \\ \mathrm{foci}\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{is}\left(0,\pm \mathrm{ae}\right)=\left(0,\pm 3\right) \\ \mathrm{For}\mathrm{hyperbola} \\ {\mathrm{e}}^2=\left(1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\right) \\ \frac{25}{9}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2} \\ 25{\mathrm{a}}^2=9{\mathrm{a}}^2+9{\mathrm{b}}^2 \\ 16{\mathrm{a}}^2=9{\mathrm{b}}^2 \\ 16{\mathrm{a}}^2-9{\mathrm{b}}^2=0...\left(1\right) \\ \mathrm{Equation}\mathrm{of}\mathrm{hyperbola}\mathrm{is} \\ \frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{As}\mathrm{it}\mathrm{passes}\mathrm{through}\left(0,\pm 3\right) \\ \frac{9}{{\mathrm{a}}^2}-0=1 \\ {\mathrm{a}}^2=9 \\ \mathrm{So}, \\ 16\left(9\right)-9{\mathrm{b}}^2=0 \\ {\mathrm{b}}^2=16 \\ \mathrm{So},\mathrm{Hyperbola}\mathrm{is} \\ \frac{{\mathrm{y}}^2}{9}-\frac{{\mathrm{x}}^2}{16}=1\mathrm{ANS}... \end{gathered}$$