Question

The equation of the hyperbola having its eccentricity 2 and the distance between foci 8, is ________________________.

Answer 1

For the hyperbola,
Eccentricity is given i.e. 2 and distance between foci is 8
$$\begin{gathered} \mathrm{i}.\mathrm{e}.2ae=8 \\ \mathrm{i}.\mathrm{e}.2a\times 2=8 \\ \mathrm{i}.\mathrm{e}.a=2 \\ \mathrm{i}.\mathrm{e}.b^2=a^2\left(-1+e^2\right)=a^2\left(e^2-1\right) \\ \mathrm{i}.\mathrm{e}.b^2=4\left(-1+4\right) \\ =4\left(+3\right) \\ {b}^2=+12 \\ \therefore \mathrm{equation}\mathrm{of}\mathrm{hyperbola}\mathrm{is}\frac{{\mathrm{x}}^2}{4}-\frac{y^2}{12}=1 \\ \mathrm{i}.\mathrm{e}.3x^2-y^2=12 \end{gathered}$$