Question

The eccentricity of the ellipse \( 9x^2+16y^2=144 \) is

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

Hint:

Standard equation of ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
If \(a^2>b^2\), major axis is along x-axis, \(a\) is semi-major axis, \(b\) is semi-minor axis. Eccentricity \(e = \sqrt{1 - b^2/a^2}\).
If \(b^2>a^2\), major axis is along y-axis, \(b\) is semi-major axis, \(a\) is semi-minor axis. Eccentricity \(e = \sqrt{1 - a^2/b^2}\).
Alternatively, \(c^2 = a^2-b^2\) (if \(a>b\)) where \(c=ae\). So \(e = c/a = \sqrt{a^2-b^2}/a\).

Answer 1

The Correct Option is C

Given: The equation of the ellipse is:

\[ 9x^2 + 16y^2 = 144 \]

Step 1: Convert to standard form

Divide both sides by 144: \[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \Rightarrow \frac{x^2}{16} + \frac{y^2}{9} = 1 \]

This matches the standard ellipse form:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad \text{with } a^2 = 16,\, b^2 = 9 \]

Since \( a^2 > b^2 \), the major axis is along the x-axis. Thus:

\[ a = 4, \quad b = 3 \]

Step 2: Calculate eccentricity

The eccentricity \( e \) of an ellipse is: \[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4} \]

✅ Final Answer:

\[ \boxed{\frac{\sqrt{7}}{4}} \]

This corresponds to option (c).