Question

The eccentricity of \((\frac {x}{25})^2 + (\frac {y}{16})^2 = 1\) is:

Note: Assuming the intended equation is \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) based on the options. The literal interpretation \( \frac{x^2}{625} + \frac{y^2}{256} = 1 \) yields\(e =\frac {\sqrt{369}}{25}\), which is not among the options.

• A. \( 3/4 \)
• B. \( 1 \)
• C. \( 3/5 \)
• D. None of these

Hint:

For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \): - If \( a > b \), the major axis is horizontal, and \( e = \sqrt{1 - b^2/a^2} \). - If \( b > a \), the major axis is vertical, and \( e = \sqrt{1 - a^2/b^2} \). Eccentricity \( e \) is always between 0 and 1 for an ellipse (\( 0 \le e < 1 \)).

Answer 1

The Correct Option is C

Step 1: Identify the standard form of the ellipse equation. The given equation is \( (x/25)^2 + (y/16)^2 = 1 \). Assuming the intended equation is \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \), it is in the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

 Step 2: Determine the values of \( a^2 \) and \( b^2 \). Comparing \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) with \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we have: \[ a^2 = 25 \quad \Rightarrow \quad a = 5 \] \[ b^2 = 16 \quad \Rightarrow \quad b = 4 \] Step 3: Identify the major axis. Since \( a^2 > b^2 \) (\( 25 > 16 \)), the major axis lies along the x-axis. Step 4: Use the formula for eccentricity (e) for an ellipse where \( a^2 > b^2 \). \[ e = \sqrt{1 - \frac{b^2}{a^2}} \] Step 5: Substitute the values of \( a^2 \) and \( b^2 \) into the formula. \[ e = \sqrt{1 - \frac{16}{25}} \] \[ e = \sqrt{\frac{25 - 16}{25}} \] \[ e = \sqrt{\frac{9}{25}} \] \[ e = \frac{\sqrt{9}}{\sqrt{25}} \] \[ e = \frac{3}{5} \] Step 6: Compare the result with the options. The calculated eccentricity is \( 3/5 \), which matches option (C).