Let L L ' be the latusrectum and S be a focus of the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` if `Delta SLL'` is equilaterial ,then the eccentricity of the ellispe ,is

Options

A. `1//sqrt(5))`
B. `1//sqrt(3)`
C. `1//sqrt(2)`
D. `sqrt(2)//3`

Answer

To solve the problem, we need to find the eccentricity of the ellipse given that triangle SLL' (where S is a focus and LL' is the latus rectum) is equilateral.

### Step-by-Step Solution:

1. **Identify the Ellipse Equation**:
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

2. **Locate the Focus and Latus Rectum**:
The coordinates of the foci of the ellipse are \((\pm ae, 0)\) where \(e\) is the eccentricity. The coordinates of the ends of the latus rectum are \((ae, \frac{b^2}{a})\) and \((ae, -\frac{b^2}{a})\).

3. **Draw the Triangle**:
We need to consider the triangle SLL' where S is the focus at \((-ae, 0)\) and L and L' are the ends of the latus rectum. Thus, the points are:
- \(S = (-ae, 0)\)
- \(L = (ae, \frac{b^2}{a})\)
- \(L' = (ae, -\frac{b^2}{a})\)

4. **Use Properties of Equilateral Triangle**:
Since triangle SLL' is equilateral, all sides are equal. We can calculate the lengths of the sides:
- Length \(SL = \sqrt{(ae + ae)^2 + \left(0 - \frac{b^2}{a}\right)^2} = \sqrt{(2ae)^2 + \left(-\frac{b^2}{a}\right)^2}\)
- Length \(SL' = \sqrt{(ae + ae)^2 + \left(0 + \frac{b^2}{a}\right)^2} = \sqrt{(2ae)^2 + \left(\frac{b^2}{a}\right)^2}\)
- Length \(LL' = \frac{b^2}{a} - \left(-\frac{b^2}{a}\right) = \frac{2b^2}{a}\)

5. **Set the Lengths Equal**:
Since \(SL = SL'\), we can set up the equation:
\[ \sqrt{(2ae)^2 + \left(\frac{b^2}{a}\right)^2} = \frac{2b^2}{a} \]

6. **Square Both Sides**:
Squaring both sides gives:
\[ (2ae)^2 + \left(\frac{b^2}{a}\right)^2 = \left(\frac{2b^2}{a}\right)^2 \]
Simplifying this:
\[ 4a^2e^2 + \frac{b^4}{a^2} = \frac{4b^4}{a^2} \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ 4a^2e^2 = \frac{4b^4}{a^2} - \frac{b^4}{a^2} = \frac{3b^4}{a^2} \]
Thus:
\[ 4a^2e^2 = \frac{3b^4}{a^2} \]

8. **Express \(b^2\) in terms of \(a\) and \(e\)**:
From the relationship \(e^2 = 1 - \frac{b^2}{a^2}\), we can express \(b^2\) as:
\[ b^2 = a^2(1 - e^2) \]

9. **Substituting \(b^2\)**:
Substitute \(b^2\) into the equation:
\[ 4a^2e^2 = \frac{3(a^2(1 - e^2))^2}{a^2} \]
Simplifying gives:
\[ 4e^2 = 3(1 - e^2)^2 \]

10. **Solve for \(e\)**:
Expanding and rearranging gives:
\[ 3(1 - 2e^2 + e^4) - 4e^2 = 0 \]
\[ 3 - 6e^2 + 3e^4 - 4e^2 = 0 \]
\[ 3e^4 - 10e^2 + 3 = 0 \]

11. **Let \(x = e^2\)**:
This is a quadratic equation in \(x\):
\[ 3x^2 - 10x + 3 = 0 \]

12. **Using the Quadratic Formula**:
The roots are given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} \]
\[ = \frac{10 \pm 8}{6} \]
Thus:
\[ x = 3 \quad \text{or} \quad x = \frac{1}{3} \]

13. **Finding \(e\)**:
Since \(x = e^2\), we have:
\[ e^2 = \frac{1}{3} \implies e = \frac{1}{\sqrt{3}} \]

### Final Answer:
The eccentricity of the ellipse is:
\[ \boxed{\frac{1}{\sqrt{3}}} \]