Key Concepts and Formulas

This problem requires us to find a common tangent to two parabolas and then calculate the perpendicular distance from a given point to this tangent line. The key concepts and formulas involved are:

1. Standard Forms of a Parabola:

   • The standard form of a parabola opening along the positive x-axis with its vertex at the origin is $y^2 = 4ax$. The focal length parameter is $a$.
   • For a parabola opening along the positive x-axis with its vertex at $(h,k)$, the standard form is $(y-k)^2 = 4a(x-h)$.

2. Equation of a Tangent to a Parabola with Slope $m$:

   • For a parabola $y^2 = 4ax$, the equation of a tangent with slope $m$ is $y = mx + \frac{a}{m}$. (This formula is valid for $m \neq 0$).
   • For a shifted parabola $(y-k)^2 = 4a(x-h)$, the equation of a tangent with slope $m$ is $(y-k) = m(x-h) + \frac{a}{m}$. (This can be derived by shifting the origin to $(h,k)$).

3. Distance from a Point to a Line: The perpendicular distance $D$ from a point $(x_1, y_1)$ to a line $Ax+By+C=0$ is given by the formula: $D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

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Step-by-Step Solution

Step 1: Analyze the First Parabola and Determine its General Tangent Equation

The first curve is given by the equation $x = 2y^2$.

• Goal: To use the standard tangent formula, we must first convert this equation into the standard form $y^2 = 4ax$.
• Action: Divide both sides by 2: $y^2 = \frac{x}{2}$
• Identification: Comparing this with the standard form $y^2 = 4ax$, we can identify the parameter $4a = \frac{1}{2}$, which implies $a = \frac{1}{8}$.
• Applying Tangent Formula: Now, we write the equation of a tangent with slope $m$ to this parabola using the formula $y = mx + \frac{a}{m}$: $y = mx + \frac{1/8}{m}$ $y = mx + \frac{1}{8m} \quad \text{(Equation 1)}$ This is the general equation of any tangent line with slope $m$ to the first parabola.

Step 2: Analyze the Second Parabola and Determine its General Tangent Equation

The second curve is given by the equation $x = 1+y^2$.

• Goal: This parabola is not centered at the origin. We need to rewrite it in the standard shifted parabola form $(y-k)^2 = 4a(x-h)$ to use the appropriate shifted tangent formula.
• Action: Rearrange the terms to isolate $y^2$: $y^2 = x-1$ This can be explicitly written as $(y-0)^2 = 1(x-1)$.
• Identification: Comparing this with $(y-k)^2 = 4a(x-h)$, we identify:
  • The vertex $(h,k) = (1,0)$.
  • The parameter $4a=1$, which implies $a = \frac{1}{4}$.
• Applying Tangent Formula: Now, we write the equation of a tangent with slope $m$ to this parabola using the formula $(y-k) = m(x-h) + \frac{a}{m}$: $y - 0 = m(x-1) + \frac{1/4}{m}$ $y = m(x-1) + \frac{1}{4m} \quad \text{(Equation 2)}$ This is the general equation of any tangent line with slope $m$ to the second parabola.

Step 3: Determine the Specific Common Tangent

• Goal: For a line to be a common tangent to both parabolas, its equation must be identical for the same slope $m$. This means that if we express both tangent equations in the form $y = mx + c$, their $y$-intercepts (the constant terms) must be equal.
• Action: From Equation 1, the $y$-intercept is $c_1 = \frac{1}{8m}$. From Equation 2, expanding $m(x-1) + \frac{1}{4m}$ gives $mx - m + \frac{1}{4m}$. So, the $y$-intercept is $c_2 = -m + \frac{1}{4m}$.
• Equating Y-intercepts: Set $c_1 = c_2$ to find the value of $m$ for the common tangent: $\frac{1}{8m} = -m + \frac{1}{4m}$
• Solving for $m$: To clear the denominators, we multiply the entire equation by $8m$. We can safely assume $m \neq 0$, because if $m=0$, the tangent would be horizontal ($y=c$), which is not possible for these parabolas that open along the x-axis. $8m \left(\frac{1}{8m}\right) = 8m \left(-m + \frac{1}{4m}\right)$ $1 = -8m^2 + \frac{8m}{4m}$ $1 = -8m^2 + 2$ Rearranging the terms to solve for $m^2$: $8m^2 = 2 - 1$ $8m^2 = 1$ $m^2 = \frac{1}{8}$ Taking the square root: $m = \pm\sqrt{\frac{1}{8}}$ $m = \pm\frac{1}{2\sqrt{2}}$
• Applying Condition: The problem statement specifies that $m > 0$. Therefore, we choose the positive value for $m$: $m = \frac{1}{2\sqrt{2}}$
• Finding the Equation of the Tangent: Now, substitute this value of $m$ back into either Equation 1 or Equation 2 to find the specific equation of the common tangent. Let's use Equation 1 for simplicity: $y = mx + \frac{1}{8m}$ Substitute $m = \frac{1}{2\sqrt{2}}$: $y = \left(\frac{1}{2\sqrt{2}}\right)x + \frac{1}{8\left(\frac{1}{2\sqrt{2}}\right)}$ $y = \frac{x}{2\sqrt{2}} + \frac{1}{\frac{4}{\sqrt{2}}}$ $y = \frac{x}{2\sqrt{2}} + \frac{\sqrt{2}}{4}$
• Converting to General Form: To get rid of the denominators and express the line in the standard general form $Ax+By+C=0$, we can multiply the entire equation by $4\sqrt{2}$ (the least common multiple of the denominators): $4\sqrt{2}y = \frac{4\sqrt{2}x}{2\sqrt{2}} + \frac{4\sqrt{2}\sqrt{2}}{4}$ $4\sqrt{2}y = 2x + 2$ Rearranging the terms to $Ax+By+C=0$: $2x - 4\sqrt{2}y + 2 = 0$ Dividing by 2 to simplify the coefficients: $x - 2\sqrt{2}y + 1 = 0 \quad \text{(Equation of the Common Tangent)}$

Step 4: Calculate the Distance from the Given Point to the Common Tangent

• Goal: Calculate the perpendicular distance from the given point $(6, -2\sqrt{2})$ to the common tangent line $x - 2\sqrt{2}y + 1 = 0$.
• Identification: From the line equation $x - 2\sqrt{2}y + 1 = 0$, we have:
  • $A = 1$
  • $B = -2\sqrt{2}$
  • $C = 1$ The given point is $(x_1, y_1) = (6, -2\sqrt{2})$.
• Applying Distance Formula: Now, we apply the distance formula $D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$: $D = \frac{|(1)(6) + (-2\sqrt{2})(-2\sqrt{2}) + 1|}{\sqrt{(1)^2 + (-2\sqrt{2})^2}}$
• Calculation of Numerator: $(1)(6) + (-2\sqrt{2})(-2\sqrt{2}) + 1 = 6 + (4 \times 2) + 1 = 6 + 8 + 1 = 15$ So, the numerator is $|15| = 15$.
• Calculation of Denominator: $\sqrt{(1)^2 + (-2\sqrt{2})^2} = \sqrt{1 + (4 \times 2)} = \sqrt{1 + 8} = \sqrt{9} = 3$
• Final Distance: Substitute these values back into the distance formula: $D = \frac{15}{3}$ $D = 5$

The distance of the point $(6, -2\sqrt{2})$ from the common tangent is 5 units.

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Tips and Common Mistakes to Avoid

• Standard Forms are Crucial: Always convert the given parabola equations into their standard forms ($y^2=4ax$, $x^2=4ay$, or their shifted versions) before applying tangent formulas. This ensures