Question

The equation of the hyperbola with vertices at (0, ± 6) and eccentricity $\frac{5}{3}$ is __________________ and its foci are __________________.

Answer 1

$$\begin{gathered} \mathrm{For}\mathrm{the}\mathrm{hyperbola}, \\ \mathrm{Vertices}\mathrm{is}\mathrm{given}\mathrm{at}\left(0,\pm 6\right)\mathrm{and}\mathrm{eccentricity}\mathrm{is}\frac{5}{3}. \\ \mathrm{Since}\mathrm{vertices}\mathrm{lie}\mathrm{on}y‐axis \\ \Rightarrow \mathrm{equation}\mathrm{of}\mathrm{hyperbola}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form},\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1 \\ \mathrm{i}.\mathrm{e}.b=6 \\ \mathrm{Since}e=\frac{5}{3} \\ \Rightarrow a^2=b^2\left(e^2-1\right) \\ \mathrm{i}.\mathrm{e}.a^2=36\left(\frac{25}{9}-1\right) \\ =36\left(\frac{16}{9}\right) \\ {a}^2=4\times 16 \\ \mathrm{i}.\mathrm{e}.\mathrm{equation}\mathrm{of}\mathrm{hyperbola}\mathrm{is}\frac{x^2}{64}-\frac{y^2}{36}=-1 \\ \mathrm{i}.\mathrm{e}.\frac{y^2}{36}-\frac{x^2}{64}=1 \\ \mathrm{and}\mathrm{foci}\mathrm{is}\left(0,\pm be\right)=\left(0,\frac{\pm 5}{3}\times 6\right)=\left(0,\pm 10\right) \end{gathered}$$