Key Concepts and Formulas for Common Tangents

To find the equation of a common tangent to two curves, we employ a systematic approach that leverages the conditions for tangency.

1. General Tangent Equation: For standard conic sections, there are often pre-derived formulas for the equation of a tangent with a given slope $m$.

   • For a parabola $y^2 = 4ax$, the equation of a tangent with slope $m$ is $y = mx + \frac{a}{m}$.
   • For a parabola $x^2 = 4ay$, the equation of a tangent with slope $m$ is $y = mx - am^2$.
   • For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$.
   • For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the equation of a tangent with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.
   • For a rectangular hyperbola $xy = c^2$, the equation of a tangent with slope $m$ is $y = mx \pm 2c\sqrt{-m}$. (Note: This is often derived from the discriminant condition or by considering parametric form. A line $y=mx+k$ is tangent to $xy=c^2$ if $k^2 = -4mc^2$).

2. Condition for Tangency (Discriminant Method): A straight line $y = mx + c$ is tangent to a quadratic curve (like a parabola, ellipse, or hyperbola) if and only if, upon substituting the line's equation into the curve's equation, the resulting quadratic equation in $x$ (or $y$) has exactly one real solution. This unique solution implies that the discriminant ($\Delta$ or $D$) of the quadratic equation must be zero. For a quadratic equation $Ax^2 + Bx + C = 0$, the discriminant is $\Delta = B^2 - 4AC$. Setting $\Delta = 0$ provides a condition for tangency.

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Step-by-Step Solution

Our goal is to find a line that is tangent to both the parabola $y^2 = 4x$ and the hyperbola $xy = 2$. The most efficient strategy is to express the general tangent for one curve (usually the simpler one) and then apply the tangency condition for the second curve.

1. Write the General Equation of a Tangent to the Parabola

The given parabola is $y^2 = 4x$.

• Explanation: This is in the standard form $y^2 = 4ax$. By comparing, we can see that $4a = 4$, which means $a = 1$. We choose to start with the parabola because its general tangent equation in terms of slope $m$ is straightforward and well-known. The general equation of a tangent to a parabola $y^2 = 4ax$ with slope $m$ is: $y = mx + \frac{a}{m}$ Substituting $a=1$ for our parabola $y^2 = 4x$: $y = mx + \frac{1}{m} \quad \text{... (1)}$
• Why this step? Equation (1) represents every possible line that is tangent to the parabola $y^2=4x$. Our task is to find the specific value(s) of $m$ for which this line is also tangent to the hyperbola $xy=2$.

2. Apply the Tangency Condition to the Hyperbola

The common tangent (represented by equation 1) must also be tangent to the hyperbola $xy = 2$.

• Explanation: For a line to be tangent to a curve, they must intersect at exactly one point. We find the intersection points by substituting the equation of the line into the equation of the curve. Substitute the expression for $y$ from the tangent line (1) into the hyperbola's equation $xy = 2$: $x \left( mx + \frac{1}{m} \right) = 2$

3. Form a Quadratic Equation in $x$

Now, we simplify and rearrange the equation obtained in Step 2 into the standard quadratic form $Ax^2 + Bx + C = 0$. $mx^2 + \frac{x}{m} = 2$ $mx^2 + \frac{1}{m}x - 2 = 0$

• Why this step? The discriminant condition for tangency ($\Delta=0$) applies directly to quadratic equations. By expressing the intersection problem as a quadratic equation in $x$, we can use this powerful tool to determine the specific slope $m$ that yields a single point of intersection (tangency). To make the coefficients integers and easier to work with, we can multiply the entire equation by $m$ (we assume $m \neq 0$, as a vertical tangent $x=constant$ cannot be tangent to $y^2=4x$ in this form): $m^2x^2 + x - 2m = 0 \quad \text{... (2)}$ From this quadratic equation, we can identify the coefficients: $A = m^2$, $B = 1$, and $C = -2m$.

4. Utilize the Discriminant Condition ($\Delta=0$)

For the line $y = mx + \frac{1}{m}$ to be tangent to the hyperbola $xy=2$, the quadratic equation (2) must have exactly one real solution for $x$. This occurs precisely when its discriminant ($\Delta$) is equal to zero.

• Explanation:
  • If $\Delta > 0$, the line intersects the curve at two distinct points (a secant).
  • If $\Delta < 0$, the line does not intersect the curve at any real point.
  • If $\Delta = 0$, the line touches the curve at exactly one point (a tangent), meaning there is one repeated real root for $x$. Recall the discriminant formula: $\Delta = B^2 - 4AC$. Using the coefficients from equation (2): $A = m^2$, $B = 1$, $C = -2m$. Set $\Delta = 0$: $(1)^2 - 4(m^2)(-2m) = 0$ $1 + 8m^3 = 0$

5. Solve for the Slope ($m$)

Now, we solve the cubic equation obtained in Step 4 for $m$: $8m^3 = -1$ $m^3 = -\frac{1}{8}$ To find $m$, we take the cube root of both sides. For a real cubic equation, there is always at least one real root. $m = \sqrt[3]{-\frac{1}{8}}$ $m = -\frac{1}{2}$

• Why this value of $m$ is unique? The equation $m^3 = -1/8$ has one real root ($m=-1/2$) and two complex conjugate roots. For a tangent line in the real coordinate plane, we are only interested in the real value of $m$. This specific value of $m$ defines the unique slope of the common tangent.

6. Construct the Equation of the Common Tangent

Substitute the value of $m = -\frac{1}{2}$ back into the general tangent equation for the parabola (equation 1): $y = \left(-\frac{1}{2}\right)x + \frac{1}{\left(-\frac{1}{2}\right)}$ $y = -\frac{1}{2}x - 2$ To express the equation in the standard form $Ax+By+C=0$ and eliminate fractions, multiply the entire equation by 2: $2y = -x - 4$ Rearrange the terms to bring them all to one side: $x + 2y + 4 = 0$

7. Compare with Options and Address Discrepancy

Let's compare our derived equation with the given options: (A) $x + y + 1 = 0$ (B) $4x + 2y + 1 = 0$ (C) $x – 2y + 4 = 0$ (D) $x + 2y + 4 = 0$

Our calculated equation, $x + 2y + 4 = 0$, matches Option (D).

Important Note on Discrepancy: The problem statement indicates "Correct Answer: A". However, our rigorous step-by-step mathematical derivation consistently leads to $x + 2y + 4 = 0$. Let's quickly verify why option (A) $x + y + 1 = 0$ would be incorrect: If the common tangent were $x+y+1=0$, its slope would be $m=-1$. The constant term is $c=1$. For this line to be tangent to $y^2=4x$, we would need $c = a/m$, so $1 = 1/(-1)$, which means $1 = -1$, which is false. Thus, $x+y+1=0$ is not tangent to the parabola $y^2=4x$. Alternatively, using our derived condition $1 + 8m^3 = 0$ (from Step 4), if $m=-1$: $1 + 8(-1)^3 = 1 - 8 = -7$. Since $-7 \neq 0$, a line with slope $m=-1$ is not tangent to the hyperbola $xy=2$. Therefore, there is a clear mismatch between the provided "Correct Answer" (A) and the result obtained through standard mathematical methods. Based on our derivation, Option (D) is the mathematically correct answer.

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Alternative Method for Hyperbola Tangency (Verification)

For a rectangular hyperbola of the form $xy=c^2$, a line $y=mx+k$ is tangent if and only if $k^2 = -4mc^2$.

• Explanation: This is a specific tangency condition for this type of hyperbola, offering a shortcut and a way to cross-verify our result obtained via the discriminant method. In our problem, the hyperbola is $xy=2$, so $c^2=2$. The tangent line (from the parabola) is $y = mx + \frac{1}{m}$. Comparing this with $y=mx+k$, we have $k = \frac{1}{m}$. Applying the tangency condition $k^2 = -4mc^2$: $\left(\frac{1}{m}\right)^2 = -4m(2)$ $\frac{1}{m^2} = -8m$ Multiply both sides by $m^2$ (assuming $m \neq 0$): $1 = -8m^3$ $8m^3 = -1$ $m^3 = -\frac{1}{8}$ $m = -\frac{1}{2}$ This confirms the value of $m$ obtained using the discriminant method, reinforcing the correctness of our solution. Substituting $m = -\frac{1}{2}$ back into $y = mx + \frac{1}{m}$ yields $x + 2y + 4 = 0$.

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Tips for Solving Similar Problems

• Know Your Standard Forms and Tangent Equations: Memorizing the general tangent equations for standard conic sections (parabola, ellipse, hyperbola, circle) can significantly save time.
• The Discriminant is a Universal Tool: For tangency problems involving lines and quadratic curves, the $\Delta=0$ condition is a fundamental, reliable, and universally applicable tool. If you don't recall a specific tangent formula, the discriminant method will always work.
• Algebraic Precision is Key: Be extremely careful with algebraic manipulations, especially signs, fractions, and exponents. A small error can lead to an incorrect value of $m$ and thus the wrong tangent equation.
• Consider Parametric Forms: For some curves (like hyperbolas), using parametric equations for a point on the curve, finding the tangent at that point, and then imposing conditions can be an alternative approach. However, for common tangents, the slope-intercept form is often more direct.
• Check Your Work (If Time Permits): If you have time, quickly substitute your final tangent equation back into both original curve equations. For a tangent, there should be exactly one point of intersection with each curve.

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Summary and Key Takeaway

To find a common tangent to two curves, a robust and generally applicable strategy involves:

1. Parameterize One Tangent: Write the general equation of a tangent to one of the curves (often the one with a simpler tangent formula) using a parameter, typically its slope $m$.
2. Impose Tangency on the Second Curve: Substitute this general tangent equation into the equation of the second curve.
3. Apply Discriminant Condition: If the resulting equation is quadratic, set its discriminant to zero ($\Delta=0$). This condition ensures that the line is tangent to the second curve.
4. Solve and Substitute: Solve the resulting equation for the parameter ($m$ in this case) and substitute it back into the general tangent equation to find the specific common tangent.

This problem effectively demonstrates the power and versatility of the discriminant condition for determining tangency between a line and a conic section. It also serves as a reminder to always trust your mathematical derivation, even if it conflicts with a provided answer option.