Key Concepts and Formulas:

This problem requires a strong understanding of tangents to conic sections, specifically parabolas and ellipses. We will utilize the following fundamental concepts:

1. Point on a Curve: If a point $(x_1, y_1)$ lies on a curve defined by an equation $f(x,y)=0$, then its coordinates must satisfy the equation, i.e., $f(x_1, y_1)=0$. This is a basic but crucial concept for establishing initial relationships.

2. Equation of Tangent to a Parabola ($y^2 = 4ax$): The equation of the tangent at a point $(x_1, y_1)$ on the parabola $y^2 = 4ax$ is given by $yy_1 = 2a(x+x_1)$. This formula allows us to directly write the tangent line's equation if we know the point of tangency and the parabola's parameter $a$.

3. Standard Form of an Ellipse: The standard equation of an ellipse centered at the origin is $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$. Here, $A^2$ and $B^2$ represent the squares of the denominators of $x^2$ and $y^2$ respectively. Identifying these values is essential for applying tangency conditions.

4. Condition of Tangency for an Ellipse: A straight line $y = mx + c$ is tangent to the ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ if and only if $c^2 = A^2m^2 + B^2$. This is a crucial condition for solving problems involving common tangents or when a line is tangent to an ellipse. It relates the slope and y-intercept of the line to the parameters of the ellipse.

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Step-by-Step Solution:

Step 1: Establish a relationship between $\alpha$ and $\beta$ using the parabola's equation.

• Concept Used: If a point lies on a curve, its coordinates satisfy the curve's equation.
• Why this step? The problem states that the point $P(\alpha, \beta)$ lies on the parabola $y^2 = x$. This initial relationship is fundamental as it connects the two variables defining the point of tangency, which will be essential for solving the problem later by reducing the number of unknowns.
• We are given the parabola $y^2 = x$ and the point $(\alpha, \beta)$ lies on it.
• Substituting the coordinates of the point into the parabola's equation: $\beta^2 = \alpha \quad \text{.........(1)}$
• We are also given that $\beta > 0$. Since $\beta^2 = \alpha$, this implies that $\alpha$ must be positive ($\alpha > 0$). This condition will be crucial for validating our final answer, as we cannot have a negative value for $\alpha$ if $\beta$ is real.

Step 2: Determine the equation of the tangent to the parabola at $(\alpha, \beta)$.

• Concept Used: Equation of tangent to a parabola $y^2 = 4ax$ at $(x_1, y_1)$ is $yy_1 = 2a(x+x_1)$.
• Why this step? The problem states that this tangent line is also a tangent to the ellipse. Therefore, we must first find the equation of this specific line using the parabola's properties. This tangent line will then be used with the ellipse's tangency condition.
• The given parabola is $y^2 = x$. To use the standard tangent formula, we compare this with the standard form $y^2 = 4ax$. From comparison, we can identify $4a = 1$, which means $a = \frac{1}{4}$.
• The point of tangency is $(x_1, y_1) = (\alpha, \beta)$.
• Substitute these values into the tangent formula: $y(\beta) = 2\left(\frac{1}{4}\right)(x+\alpha)$ $y\beta = \frac{1}{2}(x+\alpha)$
• To use the tangency condition for the ellipse, which uses the form $y = mx+c$, we need to rearrange this equation: $2y\beta = x+\alpha$ $y = \frac{1}{2\beta}x + \frac{\alpha}{2\beta}$
• From this equation, we can clearly identify the slope $m$ and the y-intercept $c$ of the tangent line: $m = \frac{1}{2\beta} \quad \text{and} \quad c = \frac{\alpha}{2\beta}$
• Tip for JEE: Always ensure your tangent line equation is in the $y = mx+c$ form before applying tangency conditions. Incorrectly identifying $m$ or $c$ due to algebraic errors or not rearranging properly is a common mistake that can lead to an incorrect final answer.

Step 3: Prepare the ellipse equation for the tangency condition.

• Concept Used: Standard form of an ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$.
• Why this step? The condition of tangency ($c^2 = A^2m^2 + B^2$) requires us to know the values of $A^2$ and $B^2$ from the ellipse's equation. We must first convert the given equation into its standard form to correctly identify these parameters.
• The given equation of the ellipse is $x^2 + 2y^2 = 1$.
• To convert this into the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, we can rewrite the coefficients: $\frac{x^2}{1} + \frac{y^2}{1/2} = 1$
• Comparing this with the standard form, we identify: $A^2 = 1 \quad \text{and} \quad B^2 = \frac{1}{2}$
• Tip for JEE: In the tangency condition $c^2 = A^2m^2 + B^2$, $A^2$ always refers to the denominator of $x^2$ and $B^2$ to the denominator of $y^2$, irrespective of whether $A^2 > B^2$ or $B^2 > A^2$. Do not confuse these with the semi-major/minor axes if the major axis is along the y-axis, as that depends on the relative values of $A^2$ and $B^2$.

Step 4: Apply the condition of tangency for the ellipse.

• Concept Used: A line $y=mx+c$ is tangent to the ellipse $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ if $c^2 = A^2m^2 + B^2$.
• Why this step? This is the crucial step that connects the tangent line (which we derived from the parabola in Step 2) to the ellipse. By substituting the expressions for $m$ and $c$ (in terms of $\alpha$ and $\beta$) and the ellipse parameters ($A^2, B^2$), we will obtain a new equation relating $\alpha$ and $\beta$.
• Substitute the values of $m = \frac{1}{2\beta}$, $c = \frac{\alpha}{2\beta}$, $A^2 = 1$, and $B^2 = \frac{1}{2}$ into the tangency condition: $\left(\frac{\alpha}{2\beta}\right)^2 = (1)\left(\frac{1}{2\beta}\right)^2 + \frac{1}{2}$
• Simplify the equation: $\frac{\alpha^2}{4\beta^2} = \frac{1}{4\beta^2} + \frac{1}{2}$
• To eliminate the denominators and simplify further, multiply the entire equation by $4\beta^2$. Note that since $\beta > 0$, $4\beta^2$ is non-zero, so this multiplication is valid. $4\beta^2 \left(\frac{\alpha^2}{4\beta^2}\right) = 4\beta^2 \left(\frac{1}{4\beta^2}\right) + 4\beta^2 \left(\frac{1}{2}\right)$ $\alpha^2 = 1 + 2\beta^2 \quad \text{.........(2)}$

Step 5: Solve for $\alpha$ using the system of equations.

• Why this step? We now have two independent equations involving $\alpha$ and $\beta$ (equation (1) from Step 1 and equation (2) from Step 4). Our ultimate goal is to find the value of $\alpha$. By substituting one equation into the other, we can eliminate $\beta$ and obtain a single equation solely in terms of $\alpha$.
• Our two key equations are:
  1. $\beta^2 = \alpha$ (from Step 1)
  2. $\alpha^2 = 1 + 2\beta^2$ (from Step 4)
• Substitute $\beta^2 = \alpha$ from equation (1) into equation (2): $\alpha^2 = 1 + 2(\alpha)$
• Rearrange this into a standard quadratic equation form $A\alpha^2+B\alpha+C=0$: $\alpha^2 - 2\alpha - 1 = 0$
• Now, solve this quadratic equation for $\alpha$ using the quadratic formula $\alpha = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$: Here, $A=1$, $B=-2$, $C=-1$. $\alpha = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$ $\alpha = \frac{2 \pm \sqrt{4 + 4}}{2}$ $\alpha = \frac{2 \pm \sqrt{8}}{2}$ $\alpha = \frac{2 \pm 2\sqrt{2}}{2}$ $\alpha = 1 \pm \sqrt{2}$
• This gives us two possible values for $\alpha$: $\alpha_1 = 1 + \sqrt{2}$ $\alpha_2 = 1 - \sqrt{2}$

Step 6: Validate the solution for $\alpha$ based on given conditions.

• Why this step? It's crucial to check if all solutions derived mathematically are consistent with the initial conditions provided in the problem statement. Failing to do so can lead to an incorrect answer, especially in multiple-choice questions where an invalid solution might be presented as an option.
• Recall from Step 1 that $\beta^2 = \alpha$.
• Since $\beta$ is a real number (as it represents a coordinate on the parabola), $\beta^2$ must be non-negative. Therefore, $\alpha$ must also be non-negative ($\alpha \ge 0$).
• Let's check our two possible values for $\alpha$:
  • For $\alpha_1 = 1 + \sqrt{2}$: Since $\sqrt{2} \approx 1.414$, $\alpha_1 \approx 1 + 1.414 = 2.414$. This value is positive, so it is a valid candidate.
  • For $\alpha_2 = 1 - \sqrt{2}$: Since $\sqrt{2} \approx 1.414$, $\alpha_2 \approx 1 - 1.414 = -0.414$. This value is negative.
• Since $\alpha$ must be non-negative ($\alpha \ge 0$), $\alpha = 1 - \sqrt{2}$ is not a valid solution.
• Therefore, the only valid value for $\alpha$ is $1 + \sqrt{2}$.

The final answer is $\alpha = 1 + \sqrt{2}$.

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Summary and Key Takeaway:

This problem is a classic example of how to tackle common tangent problems involving different conic sections. The methodical approach involves:

1. Establishing initial relationships: Use the given point of tangency to establish relationships specific to the first curve (parabola), such as $\beta^2 = \alpha$.
2. Deriving the tangent line equation: Find the equation of the tangent line to the first curve at the given point, and express it in the standard $y=mx+c$ form.
3. Standardizing the second curve's equation: Convert the equation of the second curve (ellipse) into its standard form to easily identify its parameters ($A^2, B^2$).
4. Applying the tangency condition: Use the specific condition for tangency ($c^2 = A^2m^2 + B^2$) for the second curve. This step connects the tangent line (derived from the first curve) to the second curve.
5. Solving the system of equations: Solve the resulting system of algebraic equations to find the desired variable.
6. Validating the solutions: Critically, always validate the obtained solutions against any initial conditions provided in the problem statement (like $\beta > 0$ or $\alpha \ge 0$ in this case). This step is crucial for selecting the correct answer.

This problem reinforces the importance of knowing standard forms and tangency conditions for parabolas and ellipses, along with careful algebraic manipulation and logical validation.