Question

Find the length of latus rectum, transverse axis, conjugate axis, vertices , foci, eccentricity and equations of directrices of the hyperbola 9x2-16y2=144

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{is}, \\ 9{\mathrm{x}}^2-16{\mathrm{y}}^2=144 \\ \Rightarrow \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=1 \\ \mathrm{Comparing}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{with}, \\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,\mathrm{we}\mathrm{get} \\ \mathrm{a}=4;\mathrm{b}=3 \\ \mathrm{Now},\mathrm{c}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}=\sqrt{{\left(4\right)}^2+{\left(3\right)}^2}=\sqrt{16+9}=5 \\ \mathrm{Now},\mathrm{length}\mathrm{of}\mathrm{latus}\mathrm{rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{2\times 9}{4}=\frac{9}{2}\mathrm{units} \\ \mathrm{Length}\mathrm{of}\mathrm{transverse}\mathrm{axis}=2\mathrm{a}=2\times 4=8\mathrm{units} \\ \mathrm{Length}\mathrm{of}\mathrm{conjugate}\mathrm{axis}=2\mathrm{b}=2\times 3=6\mathrm{units} \\ \mathrm{Now},\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{vertices}\mathrm{are}:\left(\pm \mathrm{a},0\right)=\left(\pm 4,0\right) \\ \mathrm{Now},\mathrm{coordinates}\mathrm{of}\mathrm{foci}\mathrm{are}::\left(\pm \mathrm{c},0\right)=\left(\pm 5,0\right) \\ \mathrm{Eccentricity},\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}=\frac{5}{4} \\ \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{directrices}\mathrm{are}:=\mathrm{x}=\pm \frac{\mathrm{a}}{\mathrm{e}}\Rightarrow \mathrm{x}=\pm \frac{4}{5/4}\Rightarrow \mathrm{x}=\pm \frac{16}{5}\Rightarrow 5\mathrm{x}\pm 16=0 \end{gathered}$$