Introduction and Problem Setup

This problem asks us to determine the value of a parameter $\alpha$ for a parabola, given that a specific line $L$ is tangent to both a hyperbola and this parabola. The core strategy to solve this involves applying the standard tangency conditions for a line to each of these conic sections.

First, let's analyze the given line $L$ and express it in the standard slope-intercept form, which is essential for applying tangency conditions.

1. Analyzing the Tangent Line L

The equation of the line $L$ is given as $2x + y = k$. To identify its slope ($m$) and y-intercept ($c$), we rearrange it into the form $y = mx + c$: $y = -2x + k$ By comparing $y = -2x + k$ with $y = mx + c$, we can clearly identify:

• The slope $m = -2$
• The y-intercept $c = k$

We are also given that $k > 0$. This condition will be important when solving for $k$.

2. Tangency Condition for the Hyperbola

The problem states that line $L$ is tangent to the hyperbola $x^2 - y^2 = 3$. We will use the standard tangency condition for a hyperbola.

Key Concept: A line $y = mx + c$ is tangent to the hyperbola given by the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if and only if the following condition holds: $c^2 = a^2m^2 - b^2$

Step-by-step Working:

• Step 1: Convert the Hyperbola Equation to Standard Form. The given hyperbola equation is $x^2 - y^2 = 3$. To match the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the right-hand side must be $1$. Therefore, we divide the entire equation by $3$: $\frac{x^2}{3} - \frac{y^2}{3} = 1$ Explanation: This step is crucial because the parameters $a^2$ and $b^2$ in the tangency condition are defined with respect to the standard form where the right-hand side is $1$. Incorrectly identifying $a^2$ and $b^2$ will lead to an incorrect result.

• Step 2: Identify Parameters $a^2$ and $b^2$. Comparing the standardized hyperbola equation $\frac{x^2}{3} - \frac{y^2}{3} = 1$ with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we can identify: $a^2 = 3 \quad \text{and} \quad b^2 = 3$

• Step 3: Apply the Tangency Condition. Now, we substitute the values of $m = -2$ (from line $L$), $c = k$ (from line $L$), $a^2 = 3$, and $b^2 = 3$ into the tangency condition $c^2 = a^2m^2 - b^2$: $k^2 = (3)(-2)^2 - 3$ $k^2 = (3)(4) - 3$ $k^2 = 12 - 3$ $k^2 = 9$ Explanation: We are using the given information that the line is tangent to the hyperbola to establish a relationship involving $k$. This allows us to find the specific value of $k$ that defines the tangent line.

• Step 4: Determine the Value of k. Taking the square root of both sides of $k^2 = 9$: $k = \pm \sqrt{9}$ $k = \pm 3$ The problem statement specifies that $k > 0$. Therefore, we must choose the positive value: $k = 3$ Explanation: The condition $k > 0$ is essential here to uniquely determine the value of $k$. This value of $k$ now precisely defines the tangent line $L$ as $y = -2x + 3$. This specific line is the one that is also tangent to the parabola.

Tip: Always ensure the hyperbola equation is in its standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$) before identifying $a^2$ and $b^2$. If the equation was $y^2 - x^2 = 3$, it would be $\frac{y^2}{3} - \frac{x^2}{3} = 1$, which would mean $a^2=3$ for the $y^2$ term and $b^2=3$ for the $x^2$ term, leading to a tangency condition $c^2 = a^2m^2 - b^2$ (if $y^2$ is the positive term) or $c^2 = b^2m^2 - a^2$ (if $x^2$ is the positive term). In our case $x^2$ is positive, so the standard form for hyperbola with transverse axis along x-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

3. Tangency Condition for the Parabola

Now we know the specific tangent line is $y = -2x + 3$. This line is also tangent to the parabola $y^2 = \alpha x$. We will use the standard tangency condition for a parabola.

Key Concept: A line $y = mx + c$ is tangent to the parabola given by the standard equation $y^2 = 4Ax$ if and only if the following condition holds: $c = \frac{A}{m}$

Step-by-step Working:

• Step 1: Convert the Parabola Equation to Standard Form. The given parabola equation is $y^2 = \alpha x$. To match the standard form $y^2 = 4Ax$, we need to express the coefficient of $x$ as $4A$. Comparing $y^2 = \alpha x$ with $y^2 = 4Ax$, we can see that: $4A = \alpha \implies A = \frac{\alpha}{4}$ Explanation: This step is crucial for correctly applying the tangency condition. The parameter $A$ in the tangency formula $c = A/m$ refers to the 'focal length' parameter in the standard form $y^2 = 4Ax$. It's a common mistake to directly use $\alpha$ instead of $A$.

• Step 2: Identify Parameters. From our tangent line $y = -2x + 3$:

  • The slope $m = -2$.
  • The y-intercept $c = 3$ (since we found $k=3$). From the parabola's standard form:
  • The parameter $A = \frac{\alpha}{4}$.

• Step 3: Apply the Tangency Condition. Now, we substitute the values of $m = -2$, $c = 3$, and $A = \frac{\alpha}{4}$ into the tangency condition $c = \frac{A}{m}$: $3 = \frac{\frac{\alpha}{4}}{-2}$ Explanation: We use the tangency condition for the parabola, along with the characteristics of the tangent line (which we found using the hyperbola's tangency), to set up an equation that allows us to solve for the unknown $\alpha$.

• Step 4: Solve for $\alpha$. Simplify the equation: $3 = \frac{\alpha}{4 \times (-2)}$ $3 = \frac{\alpha}{-8}$ To solve for $\alpha$, multiply both sides by $-8$: $\alpha = 3 \times (-8)$ $\alpha = -24$ Explanation: This final calculation directly yields the value of $\alpha$ by isolating it from the tangency equation.

Common Mistake: A frequent error is to directly use $\alpha$ as the $A$ value in the tangency condition $c = A/m$ without converting $y^2 = \alpha x$ to $y^2 = 4Ax$ first. This would lead to $3 = \frac{\alpha}{-2}$ and thus $\alpha = -6$, which is incorrect. Always ensure you correctly identify the 'focal length' parameter $A$ for the given form of the parabola.

4. Conclusion and Summary

We have successfully determined the value of $\alpha$ to be $-24$.

Summary and Key Takeaway:

This problem is a classic example of how to link different conic sections using a common tangent line. The problem-solving strategy involved:

1. Standardizing the Line Equation: Expressing $L: 2x + y = k$ as $y = -2x + k$ to clearly identify its slope ($m=-2$) and y-intercept ($c=k$).
2. Applying Hyperbola Tangency: Using the condition $c^2 = a^2m^2 - b^2$ for the hyperbola $x^2 - y^2 = 3$ (after standardizing it to $\frac{x^2}{3} - \frac{y^2}{3} = 1$) to find the value of $k$. The condition $k>0$ was crucial here, yielding $k=3$.
3. Applying Parabola Tangency: Using the condition $c = A/m$ for the parabola $y^2 = \alpha x$ (after standardizing it to $y^2 = 4Ax$, where $A = \frac{\alpha}{4}$) and the values of $m=-2$ and $c=3$ to solve for $\alpha$.

The final answer is $\boxed{\text{-24}}$.