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:

Eccentricity of an ellipse is always between 0 and 1. If you get a value greater than or equal to 1 (like option A), you have made a calculation error, likely by swapping \(a^2\) and \(b^2\) in the formula.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the equation of an ellipse and we need to calculate its eccentricity.

Step 2: Key Formula or Approach:
First, convert the equation of the ellipse to the standard form: \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\) (for a vertical ellipse) or \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (for a horizontal ellipse), where \(a \gt b\).
The eccentricity \(e\) is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Step 3: Detailed Explanation:
The given equation is \(16x^2 + 7y^2 = 112\).
To convert it to standard form, divide the entire equation by 112:
\[ \frac{16x^2}{112} + \frac{7y^2}{112} = \frac{112}{112} \]
\[ \frac{x^2}{7} + \frac{y^2}{16} = 1 \]
This is the standard form of an ellipse. We can identify \(a^2\) and \(b^2\). By convention, \(a^2\) is the larger denominator.
Here, \(a^2 = 16\) and \(b^2 = 7\).
Since \(a^2\) is under the \(y^2\) term, the major axis of the ellipse is vertical.
Now we calculate the eccentricity \(e\):
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{7}{16}} \]
\[ e = \sqrt{\frac{16-7}{16}} = \sqrt{\frac{9}{16}} \]
\[ e = \frac{3}{4} \]

Step 4: Final Answer:
The eccentricity of the ellipse is \(\frac{3}{4}\).