This problem asks us to find the slope of a line that is simultaneously tangent to two different parabolas. This means the line must satisfy the tangency conditions for both parabolas. The most efficient way to solve such problems is to use the standard equations of tangent lines in slope form for each parabola and then equate their $y$-intercepts, as the slope $m$ will be common to both.

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1. Key Concepts: Tangent Equations in Slope Form

For a line $y = mx + c$ to be tangent to a parabola, there's a specific relationship between its $y$-intercept $c$, its slope $m$, and the parabola's characteristic parameter 'a'. Understanding these standard forms is crucial for solving problems involving tangents.

• Tangent to a Parabola of the form $y^2 = 4ax$: The equation of a tangent line with slope $m$ to the parabola $y^2 = 4ax$ is given by: $y = mx + \frac{a}{m}$

  • Explanation: This formula is derived by substituting $y = mx+c$ into the parabola's equation, which gives $(mx+c)^2 = 4ax$. Expanding this results in a quadratic equation in $x$: $m^2x^2 + (2mc - 4a)x + c^2 = 0$. For the line to be tangent, it must intersect the parabola at exactly one point, meaning this quadratic equation must have exactly one solution. This occurs when its discriminant ($D$) is zero. Setting $D = (2mc - 4a)^2 - 4m^2c^2 = 0$ and solving for $c$ in terms of $a$ and $m$ yields $c = a/m$.
  • Note: This formula is valid for all finite slopes $m \neq 0$. The vertical tangent ($x=0$) has an undefined slope.

• Tangent to a Parabola of the form $x^2 = 4ay$: The equation of a tangent line with slope $m$ to the parabola $x^2 = 4ay$ is given by: $y = mx - am^2$

  • Explanation: Similarly, we substitute $y = mx+c$ into the parabola's equation: $x^2 = 4a(mx+c)$. This gives a quadratic equation in $x$: $x^2 - 4amx - 4ac = 0$. For tangency, the discriminant must be zero: $D = (-4am)^2 - 4(1)(-4ac) = 16a^2m^2 + 16ac = 0$. Dividing by $16a$ (assuming $a \neq 0$) gives $am^2 + c = 0$, which implies $c = -am^2$.
  • Tip: These two formulas are fundamental and appear frequently in JEE problems. It's highly recommended to memorize them or be able to derive them quickly.

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2. Step-by-Step Solution

Step 2.1: Determine the Tangent Equation for the First Parabola ($y^2 = 4x$)

The first parabola is given by ${y^2} = 4x$.

• Identify the Standard Form: We compare this equation with the standard form $y^2 = 4ax$.
• Determine the Parameter 'a': By direct comparison, we see that $4a = 4$. Therefore, the parameter $a = 1$.
• Formulate the Tangent Equation: Substitute $a = 1$ into the tangent formula $y = mx + \frac{a}{m}$. The equation of any line tangent to $y^2 = 4x$ with slope $m$ is: $y = mx + \frac{1}{m} \quad \cdots (1)$
  • Rationale: This equation represents all possible non-vertical tangent lines to the first parabola. We need to find the specific line that also touches the second parabola, which means its slope $m$ and $y$-intercept $c$ must be the same for both.

Step 2.2: Determine the Tangent Equation for the Second Parabola ($x^2 = -32y$)

The second parabola is given by ${x^2} = - 32y$.

• Identify the Standard Form: We compare this equation with the standard form $x^2 = 4ay$.
• Determine the Parameter 'a': By direct comparison, we see that $4a = -32$. Therefore, the parameter $a = -8$.
  • Common Mistake Alert: It is crucial to correctly identify the sign of 'a'. Here, 'a' is negative, indicating that the parabola opens downwards. A common error is to mistakenly use $a=8$.
• Formulate the Tangent Equation: Substitute $a = -8$ into the tangent formula $y = mx - am^2$. The equation of any line tangent to $x^2 = -32y$ with slope $m$ is: $y = mx - (-8)m^2$ $y = mx + 8m^2 \quad \cdots (2)$
  • Rationale: Similar to Step 2.1, this equation represents all possible tangent lines to the second parabola. For a common tangent, this equation must be identical to Equation (1).

Step 2.3: Equate the Tangent Equations to Find the Common Slope

Since the line is tangent to both parabolas, it must be the same line. This implies that its slope $m$ and its $y$-intercept $c$ must be identical for both equations.

• Equate the y-intercepts: From Equation (1), the $y$-intercept is $c_1 = \frac{1}{m}$. From Equation (2), the $y$-intercept is $c_2 = 8m^2$. For the lines to be identical, $c_1 = c_2$. $\frac{1}{m} = 8m^2$
• Solve for m: Multiply both sides by $m$ (we can assume $m \neq 0$, because if $m=0$, then $1/m$ is undefined, and $y=0$ is not a tangent to $y^2=4x$ in a way that matches $y=8m^2$ which would be $y=0$ for $m=0$ but $y=0$ is $x$-axis, not a tangent). $1 = 8m^3$ $m^3 = \frac{1}{8}$ Take the cube root of both sides: $m = \sqrt[3]{\frac{1}{8}}$ $m = \frac{1}{2}$
  • Rationale: By equating the $y$-intercepts (or constants) of the two general tangent equations, we are forcing them to represent the same line. This allows us to solve for the unique common slope $m$.

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3. Conclusion and Key Takeaway

The slope of the line touching both parabolas is $m = \frac{1}{2}$.

The final answer is $\boxed{\text{1/2}}$.

Key Takeaway: This problem beautifully illustrates the power of using standard tangent equations in slope form. The core strategy involves:

1. Identifying the correct standard form for each parabola.
2. Accurately determining the parameter 'a' for each.
3. Applying the corresponding tangent formula to get an equation in terms of $m$.
4. Equating the $y$-intercepts (constant terms) of these two equations to solve for the common slope $m$. Always double-check the signs of 'a' and the formulas themselves to avoid common algebraic errors.