Question

The eccentricity of the ellipse \(16x^2+7y^2=112\) is

• A. \(\frac{4}{3}\)
• B. \(\frac{7}{16}\)
• C. \(\frac{3}{\sqrt{7}}\)
• D. \(\frac{3}{4}\)

Hint:

For an ellipse, first convert the equation into standard form. The larger denominator is \(a^2\), and eccentricity is \(e=\sqrt{1-\frac{b^2}{a^2}}\).

Answer 1

The Correct Option is D

We are given the ellipse: \[ 16x^2+7y^2=112. \] First convert it into standard form. Divide both sides by \(112\): \[ \frac{16x^2}{112}+\frac{7y^2}{112}=1. \] \[ \frac{x^2}{7}+\frac{y^2}{16}=1. \] The larger denominator is \(16\), so the major axis is along the \(y\)-axis. Thus, \[ a^2=16 \] and \[ b^2=7. \] Now eccentricity of ellipse is given by \[ e=\sqrt{1-\frac{b^2}{a^2}}. \] Substitute the values: \[ e=\sqrt{1-\frac{7}{16}}. \] \[ e=\sqrt{\frac{16-7}{16}}. \] \[ e=\sqrt{\frac{9}{16}}. \] \[ e=\frac{3}{4}. \] Therefore, the eccentricity of the ellipse is \[ \frac{3}{4}. \]