Question

The distance between the foci of the ellipse 3x² + 4y² = 48 is ___________.

Answer 1

Given ellipse is 3x² + 4y² = 48
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{3x^2}{48}+\frac{4y^2}{48}=1 \\ \mathrm{i}.\mathrm{e}\frac{x^2}{16}+\frac{y^2}{12}=1 \\ \mathrm{i}.\mathrm{e}{a}^2=16,b^2=12 \\ \therefore e=\sqrt{1-\frac{b^2}{a^2}} \\ =\sqrt{1-\frac{12}{16}} \\ =\sqrt{\frac{16-12}{16}} \\ =\sqrt{\frac{4}{16}} \\ \mathrm{i}.\mathrm{e}e=\frac{1}{2} \\ \therefore \mathrm{Coordinate}\mathrm{of}\mathrm{foci}\mathrm{are}\left(\pm ae,0\right) \\ \mathrm{i}.\mathrm{e}\left(\pm 4\times \frac{1}{2},0\right) \\ \mathrm{i}.\mathrm{e}\left(\pm 2,0\right) \end{gathered}$$
∴ Distance of foci is 4