Let there are exactly two points on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` whose distance from (0, 0) are equal to `sqrt((a^(2))/(2)+b^(2))`. Then, the eccentricity of the ellipse is equal to

Options

A. `(1)/(2)`
B. `(1)/(2sqrt2)`
C. `(1)/(sqrt2)`
D. None of these

Answer

To solve the problem, we need to find the eccentricity of the ellipse given that there are exactly two points on the ellipse whose distance from the origin is equal to \(\sqrt{\frac{a^2}{2} + b^2}\).

### Step-by-Step Solution:

1. **Understanding the Ellipse Equation**:
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Here, \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

2. **Distance from the Origin**:
We need to find points \((x, y)\) on the ellipse such that their distance from the origin is:
\[ \sqrt{\frac{a^2}{2} + b^2} \]
The distance from the origin to a point \((x, y)\) is given by:
\[ d = \sqrt{x^2 + y^2} \]

3. **Setting Up the Equation**:
We equate the distance to the expression given:
\[ \sqrt{x^2 + y^2} = \sqrt{\frac{a^2}{2} + b^2} \]
Squaring both sides gives:
\[ x^2 + y^2 = \frac{a^2}{2} + b^2 \]

4. **Substituting for \(y^2\)**:
From the ellipse equation, we can express \(y^2\) in terms of \(x^2\):
\[ y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right) \]
Substituting this into our distance equation:
\[ x^2 + b^2 \left(1 - \frac{x^2}{a^2}\right) = \frac{a^2}{2} + b^2 \]
Simplifying this gives:
\[ x^2 + b^2 - \frac{b^2 x^2}{a^2} = \frac{a^2}{2} + b^2 \]
Canceling \(b^2\) from both sides:
\[ x^2 - \frac{b^2 x^2}{a^2} = \frac{a^2}{2} \]

5. **Factoring Out \(x^2\)**:
Factor out \(x^2\):
\[ x^2 \left(1 - \frac{b^2}{a^2}\right) = \frac{a^2}{2} \]
Thus:
\[ x^2 = \frac{\frac{a^2}{2}}{1 - \frac{b^2}{a^2}} = \frac{a^2/2}{\frac{a^2 - b^2}{a^2}} = \frac{a^2}{2} \cdot \frac{a^2}{a^2 - b^2} \]

6. **Condition for Exactly Two Points**:
For there to be exactly two points, the expression for \(x^2\) must yield a unique solution. This occurs when:
\[ a^2 > b^2 \]
This means the ellipse is elongated along the x-axis.

7. **Finding Eccentricity**:
The eccentricity \(e\) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
From our earlier work, we found that:
\[ \frac{a^2}{b^2} = 2 \implies \frac{b^2}{a^2} = \frac{1}{2} \]
Therefore:
\[ e = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer:
The eccentricity of the ellipse is:
\[ \frac{1}{\sqrt{2}} \]