Key Concepts and Tangency Conditions for Parabolas

This problem requires us to apply the conditions for a line to be tangent to a parabola. Understanding these conditions is crucial.

1. Definition of a Tangent Line: A line is tangent to a curve if it touches the curve at exactly one point without crossing it at that point.

2. Standard Tangency Conditions for Parabolas:

   • For a parabola of the form $y^2 = 4ax$: A line $y = mx + c$ is tangent to this parabola if and only if $c = \frac{a}{m}$. The point of tangency is $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$.
   • For a parabola of the form $x^2 = 4ay$: A line $y = mx + c$ is tangent to this parabola if and only if $c = -am^2$. The point of tangency is $\left(2am, -am^2\right)$.
   • Important Note: The parameter '$a$' in $y^2=4ax$ and $x^2=4ay$ represents different focal lengths for different parabolas. When dealing with multiple parabolas, it's good practice to use distinct notations (e.g., $a_1, a_2$) if there's a chance of confusion.

3. General Tangency Condition (Discriminant Method): If you substitute the equation of a line ($y = mx + c$) into the equation of a curve, you will obtain an equation in a single variable (e.g., $x$ or $y$). If the line is tangent to the curve, this resulting equation must have exactly one solution (a repeated root). For a quadratic equation $Ax^2 + Bx + C = 0$, this means its discriminant must be zero: $D = B^2 - 4AC = 0$. This method is universally applicable and serves as a reliable fallback if specific tangency conditions are forgotten or if the parabola's axis is not parallel to the coordinate axes.

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Problem Setup and Goal

We are given a line with the equation $y = mx + 4$. This line is stated to be tangent to two different parabolas:

• Parabola 1: $y^2 = 4x$
• Parabola 2: $x^2 = 2by$

Our objective is to determine the value of the unknown parameter $b$. Since the line is tangent to both parabolas, it implies that the slope $m$ and the y-intercept $c=4$ are common to both tangency conditions. We will use this commonality to find $m$ first, and then $b$.

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Step-by-Step Solution

Step 1: Determine the slope ($m$) of the tangent line using Parabola 1 ($y^2 = 4x$).

• Why this step? The equation of the tangent line $y = mx + 4$ has an unknown slope $m$. Parabola 1 ($y^2 = 4x$) is fully defined. By applying the tangency condition between the line and Parabola 1, we can uniquely determine the value of $m$.

• Identify parameters for Parabola 1: The standard form of a parabola opening to the right is $y^2 = 4ax$. Comparing $y^2 = 4x$ with $y^2 = 4ax$, we can identify the parameter $a$. We have $4a = 4$, which implies $a = 1$.

• Identify parameters for the tangent line: The given line equation is $y = mx + 4$. Comparing this with the standard line equation $y = mx + c$, we have the y-intercept $c = 4$.

• Apply the specific tangency condition for $y^2 = 4ax$: For a line $y = mx + c$ to be tangent to $y^2 = 4ax$, the condition is $c = \frac{a}{m}$.

• Substitute known values and solve for $m$: Substitute $c=4$ and $a=1$ into the tangency condition: $4 = \frac{1}{m}$ Solving for $m$: $m = \frac{1}{4}$ Now we know the complete equation of the common tangent line: $y = \frac{1}{4}x + 4$.

Step 2: Determine the value of $b$ using Parabola 2 ($x^2 = 2by$) and the determined slope ($m$).

• Why this step? We have now fully determined the equation of the tangent line ($y = \frac{1}{4}x + 4$). Since this line is also tangent to the second parabola $x^2 = 2by$, we can use this fact to find the unknown parameter $b$. We will use the general discriminant method, which is robust and applicable to any quadratic curve.

• Substitute the line equation into Parabola 2: The equation of Parabola 2 is $x^2 = 2by$. The equation of the tangent line is $y = \frac{1}{4}x + 4$. Substitute the expression for $y$ from the tangent line equation into the parabola equation: $x^2 = 2b \left( \frac{1}{4}x + 4 \right)$

• Simplify and rearrange into a standard quadratic equation: First, distribute $2b$ on the right side: $x^2 = \frac{2b}{4}x + 8b$ $x^2 = \frac{b}{2}x + 8b$ Now, move all terms to the left side to form a standard quadratic equation $Ax^2 + Bx + C = 0$: $x^2 - \frac{b}{2}x - 8b = 0$ From this quadratic equation, we identify the coefficients: $A=1$, $B=-\frac{b}{2}$, and $C=-8b$.

• Apply the discriminant condition for tangency: For the line to be tangent to the parabola, the quadratic equation must have exactly one real root. This means its discriminant ($D = B^2 - 4AC$) must be equal to zero. $D = \left(-\frac{b}{2}\right)^2 - 4(1)(-8b) = 0$

• Solve for $b$: Calculate the square of $-\frac{b}{2}$ and multiply the other terms: $\frac{b^2}{4} + 32b = 0$ To solve for $b$, factor out $b$ from the equation: $b \left( \frac{b}{4} + 32 \right) = 0$ This equation yields two possible solutions for $b$:

  1. $b = 0$
  2. $\frac{b}{4} + 32 = 0 \Rightarrow \frac{b}{4} = -32 \Rightarrow b = -128$

Step 3: Validate the solutions for $b$.

• Why this step? It is crucial to check if all algebraic solutions are valid in the context of the original geometric problem. Sometimes, solutions might lead to degenerate cases that do not represent a true parabola.

• Check $b=0$: If $b=0$, the equation of the second parabola $x^2 = 2by$ becomes $x^2 = 2(0)y$, which simplifies to $x^2 = 0$, or simply $x=0$. The equation $x=0$ represents the y-axis. The y-axis is a degenerate case and not a parabola in the standard sense (it's a pair of coincident lines). A line cannot be "tangent" to the y-axis in the way it is tangent to a parabola defined by $x^2=2by$ where $b \neq 0$. Therefore, $b=0$ is not a valid solution in this context.

• Check $b=-128$: If $b=-128$, the equation of the second parabola is $x^2 = 2(-128)y$, which simplifies to $x^2 = -256y$. This is a valid parabola that opens downwards. Thus, $b = -128$ is the correct and only valid solution.

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Common Mistakes and Tips

• Memorize Specific Tangency Conditions: While the discriminant method is universal, knowing the specific tangency conditions ($c = a/m$ for $y^2=4ax$ and $c = -am^2$ for $x^2=4ay$) can significantly speed up problem-solving on exams.

  • Alternative for Step 2 using specific formula: For $x^2 = 2by$, we first identify $a'$ (the parameter for this parabola). Comparing $x^2 = 2by$ with $x^2 = 4a'y$, we get $4a' = 2b$, so $a' = \frac{b}{2}$. The tangent line is $y = \frac{1}{4}x + 4$, so $m=\frac{1}{4}$ and $c=4$. Applying the tangency condition $c = -a'm^2$: $4 = -\left(\frac{b}{2}\right)\left(\frac{1}{4}\right)^2$ $4 = -\left(\frac{b}{2}\right)\left(\frac{1}{16}\right)$ $4 = -\frac{b}{32}$ $b = -128$ This confirms the result obtained via the discriminant method.

• Correctly Identify Parameters: Be extremely careful when identifying $a$, $m$, and $c$ from the given equations. For example, in $y^2 = 4x$, $4a=4 \Rightarrow a=1$, not $a=4$. Similarly, for $x^2=2by$, the coefficient of $y$ is $2b$, so the parameter $a'$ (for $x^2=4a'y$) is $b/2$.

• Algebraic Precision: Pay close attention to signs and fractions throughout your calculations, especially when squaring terms and applying the discriminant formula. A small error can lead to an incorrect answer.

• Consider All Solutions and Validate: When solving quadratic or higher-order equations, always list all possible roots. Then, critically evaluate each root in the context of the original problem to discard any invalid or degenerate solutions. This is a crucial step for geometric problems.

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Summary/Key Takeaway

This problem is a classic application of parabola tangency conditions. The strategy involved:

1. Using the fully defined parabola ($y^2=4x$) and the known y-intercept ($c=4$) of the tangent line to determine its slope ($m=\frac{1}{4}$).
2. Using the complete tangent line equation ($y=\frac{1}{4}x+4$) and the second parabola ($x^2=2by$) to solve for the unknown parameter ($b=-128$). This can be done either by using the specific tangency formula for $x^2=4ay$ or by the general discriminant method.
3. Always validating the obtained algebraic solutions to ensure they represent a geometrically meaningful scenario.

The final answer is $\boxed{-128}$.