Question

If the distance between the foci is equal to the length of the latus rectum, then the eccentricity of the ellipse is

• A. \( \frac{\sqrt{5}+1}{2} \)
• B. \( \frac{1-\sqrt{5}}{2} \)
• C. \( \frac{\sqrt{5}-1}{2} \)
• D. \( \frac{1+\sqrt{5}}{2} \)

Hint:

For ellipse problems, always relate \(a, b, c\) using \(c^2=a^2-b^2\) and express everything in terms of eccentricity \(e=\frac{c}{a}\).

Answer 1

The Correct Option is C

Step 1: Recall standard ellipse parameters.
For an ellipse:
\[ \text{Distance between foci} = 2c \]
\[ \text{Length of latus rectum} = \frac{2b^2}{a} \]
Also,
\[ c^2 = a^2 - b^2,\quad e = \frac{c}{a}. \]

Step 2: Use given condition.
\[ 2c = \frac{2b^2}{a}. \]
Cancel 2:
\[ c = \frac{b^2}{a}. \]

Step 3: Substitute \( b^2 = a^2 - c^2 \).
\[ c = \frac{a^2 - c^2}{a}. \]

Step 4: Multiply by \(a\).
\[ ac = a^2 - c^2. \]
\[ c^2 + ac - a^2 = 0. \]

Step 5: Divide by \(a^2\).
Let \(e = \frac{c}{a}\), then:
\[ e^2 + e - 1 = 0. \]

Step 6: Solve quadratic equation.
\[ e = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}. \]

Step 7: Choose valid value.
Since eccentricity is positive and less than 1:
\[ e = \frac{\sqrt{5}-1}{2}. \]
Final Answer:
\[ \boxed{\frac{\sqrt{5}-1}{2}} \]