Question

The eccentricity of the ellipse 16x² + 7y² = 112 is ___________.

Answer 1

For given ellipse 16x² + 7y² =112
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{16}{112}{x}^2+\frac{7}{112}{y}^2=1 \\ \mathrm{i}.\mathrm{e}\frac{x^2}{7}+\frac{y^2}{16}=1 \\ \mathrm{i}.\mathrm{e}\frac{x^2}{{\left(\sqrt{7}\right)}^2}+\frac{y^2}{{\left(4\right)}^2}=1 \\ \mathrm{i}.\mathrm{e}a=\sqrt{7}\mathrm{and}b=4 \\ \mathrm{i}.\mathrm{e}b>a \\ \therefore \mathrm{eccentricity}\mathrm{is}\sqrt{1-\frac{a^2}{b^2}} \\ =\sqrt{1-\frac{7}{16}} \\ =\sqrt{\frac{16-7}{16}}=\sqrt{\frac{9}{16}} \\ \mathrm{i}.\mathrm{e}\mathrm{eccentricity}=\frac{3}{4} \end{gathered}$$