1. Understanding the Problem and Key Concepts

The problem asks us to find the equation of a line that is simultaneously tangent to two different curves: a parabola and a rectangular hyperbola. Such a line is called a common tangent. To solve this, we will employ a powerful and systematic approach that combines the general form of a tangent to one curve with the tangency condition for the second curve.

The core mathematical concepts we will utilize are:

• Equation of Tangent to a Parabola: For a parabola of the standard form $y^2 = 4ax$, the equation of a tangent line with slope $m$ is given by $y = mx + \frac{a}{m}$. This formula is incredibly useful because it allows us to represent any tangent to the parabola using a single unknown parameter, its slope $m$.
• Condition for Tangency ($D=0$): A line $y = mx + c$ is tangent to a curve if, when we substitute the line's equation into the curve's equation, the resulting polynomial equation has exactly one unique solution for the variable (e.g., $x$ or $y$). For a quadratic equation of the form $Ax^2 + Bx + C = 0$, this condition is met when its discriminant $D = B^2 - 4AC$ is equal to zero. This ensures that the line intersects the curve at precisely one point, which is the defining characteristic of a tangent.

Our strategy will be to first write the general equation of a tangent to the parabola in terms of its slope $m$. Then, we will substitute this general tangent into the hyperbola's equation and use the discriminant condition to find the specific value of $m$ that makes it tangent to the hyperbola as well.

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2. Step-by-Step Solution

Step 2.1: Determine the general equation of a tangent to the parabola $y^2 = 16x$.

• Explanation: We begin by finding a general algebraic expression for any line that is tangent to the given parabola. The slope-intercept form $y = mx + c$ is ideal here, as the special formula for a parabola's tangent directly gives us $c$ in terms of $m$ and the parabola's parameter $a$. This reduces the problem to finding a single unknown, $m$.

• Working: The given parabola is $y^2 = 16x$. We compare this with the standard form of a parabola $y^2 = 4ax$. By comparing the coefficients of $x$, we have $4a = 16$. Solving for $a$, we find $a = 4$.

  Now, we substitute this value of $a$ into the general tangent equation for a parabola, $y = mx + \frac{a}{m}$: $y = mx + \frac{4}{m} \quad \text{......(1)}$ This equation represents any line that is tangent to the parabola $y^2 = 16x$. Our goal is to find the specific value of $m$ for which this line is also tangent to the hyperbola.

Step 2.2: Apply the tangency condition ($D=0$) to the hyperbola $xy = -4$.

• Explanation: For the line represented by equation (1) to be a common tangent, it must not only touch the parabola but also be tangent to the hyperbola $xy = -4$. To impose this condition, we substitute the expression for $y$ from our general tangent equation (1) into the hyperbola's equation. If the line is indeed tangent, the resulting equation (which will be a quadratic in $x$) must have exactly one unique solution for $x$. This is mathematically expressed by setting its discriminant $D$ to zero.

• Working: Substitute $y = mx + \frac{4}{m}$ from equation (1) into the hyperbola equation $xy = -4$: $x \left( mx + \frac{4}{m} \right) = -4$ Distribute $x$ into the parenthesis: $mx^2 + \frac{4}{m}x = -4$ To apply the discriminant condition, we must rearrange this into the standard quadratic form $Ax^2 + Bx + C = 0$: $mx^2 + \left(\frac{4}{m}\right)x + 4 = 0 \quad \text{......(2)}$ For this quadratic equation to have exactly one solution for $x$ (which means the line is tangent to the hyperbola), its discriminant ($D$) must be zero. From equation (2), we identify the coefficients: $A = m$ $B = \frac{4}{m}$ $C = 4$ Now, we set the discriminant $D = B^2 - 4AC$ to zero: $\left(\frac{4}{m}\right)^2 - 4(m)(4) = 0$

Step 2.3: Solve for the slope ($m$).

• Explanation: We now have an algebraic equation that depends solely on the unknown slope $m$. Solving this equation will yield the specific value(s) of $m$ for which the line is tangent to both curves simultaneously. We are looking for real values of $m$, as they correspond to real, geometrically observable tangent lines.

• Working: Simplify the equation from Step 2.2: $\frac{16}{m^2} - 16m = 0$ To simplify, we can divide the entire equation by 16. It's important to note that $m \neq 0$, because if $m=0$, the term $\frac{4}{m}$ in the tangent equation (1) would be undefined. A line with $m=0$ is $y=4$ (if we consider the limit), which is not tangent to $xy=-4$. $\frac{1}{m^2} - m = 0$ Multiply the entire equation by $m^2$ to clear the denominator. Since $m \neq 0$, $m^2 \neq 0$, so this is a valid operation: $1 - m^3 = 0$ $m^3 = 1$ For a real tangent line, we are interested in real values of $m$. The only real solution for $m$ is: $m = 1$ (While $m^3=1$ has two complex roots, $m = e^{i2\pi/3}$ and $m = e^{i4\pi/3}$, these correspond to complex tangent lines, which are not typically sought in such problems unless explicitly stated.)

Step 2.4: Formulate the equation of the common tangent.

• Explanation: With the value of the slope $m$ now precisely determined, we can substitute it back into our general tangent equation (1). This will give us the specific equation of the common tangent line that satisfies the conditions for both the parabola and the hyperbola.

• Working: Substitute $m=1$ into equation (1): $y = (1)x + \frac{4}{1}$ $y = x + 4$ To match the standard format of the given options (usually $Ax + By + C = 0$), we rearrange the equation: $x - y + 4 = 0$

• Result: The equation of the common tangent to the curves $y^2 = 16x$ and $xy = -4$ is $\boxed{x - y + 4 = 0}$.

This corresponds to option (A).

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3. Important Tips and Common Pitfalls

• Correctly Identify 'a' for Parabola: Always ensure you correctly extract the value of '$a$' from the parabola's equation (e.g., $y^2 = 4ax$ or $x^2 = 4ay$) before using the tangent formula. A common mistake is to use $4a$ instead of $a$. Here, $y^2=16x \Rightarrow 4a=16 \Rightarrow a=4$.
• Discriminant Condition is Key: The $D=0$ condition is fundamental for establishing tangency. Carefully form the quadratic equation after substitution and accurately identify its coefficients ($A, B, C$) before applying $B^2 - 4AC = 0$. Any algebraic error here will lead to an incorrect value of $m$.
• Real Slopes for Real Tangents: When solving for $m$, especially if you encounter higher-degree equations (like $m^3=1$), remember that for a physically observable, real tangent line, you are looking for real values of $m$.
• Consider Special Cases (Vertical/Horizontal Tangents): The slope form $y = mx + a/m$ does not cover vertical tangents (where $m$ is undefined). Always briefly consider if such tangents exist and if they could be relevant. For $y^2 = 16x$, the tangent at the vertex $(0,0)$ is $x=0$. However, $x=0$ does not intersect $xy=-4$ at a single point (it doesn't intersect it at all, as $0 \cdot y = -4$ has no solution). For the hyperbola $xy=-4$, neither $x=k$ nor $y=k$ for any constant $k$ can be tangents in the sense of the $D=0$ condition (they intersect at one point if $k \neq 0$, but don't lead to a quadratic in $x$ or $y$ that satisfies $D=0$). In this problem, $m=1$ implies a non-vertical, non-horizontal tangent, so the formula is perfectly applicable.
• Algebraic Precision: Be meticulous with algebraic manipulations, especially when clearing denominators or solving cubic equations. A small error can lead to a completely different result.

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4. Conclusion and Key Takeaway

To find a common tangent to two curves, a highly effective and standard strategy is:

1. Parameterize one tangent: Derive the general equation of a tangent for one curve (often using the slope form $y = mx + c$, where $c$ is expressed in terms of $m$ and curve parameters).
2. Substitute into the second curve: Substitute this general tangent equation into the equation of the second curve.
3. Apply tangency condition: Use the condition for tangency (discriminant $D=0$ for quadratics, or unique solution for higher-degree polynomials) to the resulting equation.
4. Solve for the parameter: Solve the algebraic equation for the unknown parameter (typically the slope $m$).
5. Formulate the final equation: Substitute the found parameter back into the general tangent equation to obtain the specific common tangent.

This systematic method reduces the geometric problem of finding a common tangent to solving a single algebraic equation for the tangent's slope, making it a robust approach for such problems.