If tangent of `y^(2)=x` at `(alpha,beta)`, where `betagt0` is also a tangent of ellipse `x^(1)2y^(2)=1` then value of `alpha` is

Options

A. `2 sqrt(2) + 1`
B. `sqrt(2) + 1`
C. `2 sqrt(2) - 1`
D. `sqrt(2) - 1`

Explanation

To solve the problem, we need to find the value of \( \alpha \) given that the tangent to the parabola \( y^2 = x \) at the point \( (\alpha, \beta) \) (where \( \beta > 0 \)) is also a tangent to the ellipse \( x^2 + 2y^2 = 1 \).

### Step 1: Identify the point on the parabola
The point \( (\alpha, \beta) \) lies on the parabola \( y^2 = x \). Therefore, we can express \( \alpha \) in terms of \( \beta \):
\[ \beta^2 = \alpha \]

### Step 2: Write the equation of the tangent to the parabola
The equation of the tangent to the parabola \( y^2 = x \) at the point \( (\alpha, \beta) \) is given by:
\[ yy_1 = x + x_1 \]
Substituting \( y_1 = \beta \) and \( x_1 = \alpha \), we get:
\[ y\beta = x + \alpha \]
Rearranging gives:
\[ y = \frac{x}{\beta} + \alpha \]

### Step 3: Rewrite the tangent equation
From the previous step, we can express the tangent line in slope-intercept form:
\[ y = \frac{1}{\beta}x + \alpha - \frac{\alpha}{\beta} \]

### Step 4: Identify the ellipse equation
The equation of the ellipse is:
\[ x^2 + 2y^2 = 1 \]
We can rewrite this as:
\[ \frac{x^2}{1} + \frac{y^2}{\frac{1}{2}} = 1 \]
where \( a^2 = 1 \) and \( b^2 = \frac{1}{2} \).

### Step 5: Use the condition for tangency
For the line \( y = mx + c \) to be tangent to the ellipse, the condition is:
\[ c^2 = a^2 m^2 + b^2 \]
Here, \( m = \frac{1}{\beta} \) and \( c = \alpha - \frac{\alpha}{\beta} = \alpha(1 - \frac{1}{\beta}) \).

### Step 6: Substitute into the tangency condition
Substituting \( c \) and \( m \) into the tangency condition:
\[ \left(\alpha(1 - \frac{1}{\beta})\right)^2 = 1 \cdot \left(\frac{1}{\beta}\right)^2 + \frac{1}{2} \]
Expanding gives:
\[ \alpha^2(1 - \frac{2}{\beta} + \frac{1}{\beta^2}) = \frac{1}{\beta^2} + \frac{1}{2} \]

### Step 7: Simplify and solve for \( \alpha \)
Substituting \( \alpha = \beta^2 \):
\[ \beta^4(1 - \frac{2}{\beta} + \frac{1}{\beta^2}) = \frac{1}{\beta^2} + \frac{1}{2} \]
This leads to a quadratic equation in terms of \( \beta^2 \).

### Step 8: Solve the quadratic equation
After simplifying, we find:
\[ \beta^4 - 2\beta^2 - 1 = 0 \]
Using the quadratic formula:
\[ \beta^2 = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2} \]
Since \( \beta^2 > 0 \), we take:
\[ \beta^2 = 1 + \sqrt{2} \]

### Step 9: Find \( \alpha \)
Now substituting back to find \( \alpha \):
\[ \alpha = \beta^2 = 1 + \sqrt{2} \]

### Final Answer
Thus, the value of \( \alpha \) is:
\[ \boxed{1 + \sqrt{2}} \]