This problem requires us to find a common tangent to two parabolas, given their geometric properties. We will use the standard forms of parabola equations, the concept of latus rectum, and the condition for a line to be tangent to a parabola.

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1. Understanding the Problem and Key Concepts

We are given two parabolas with specific characteristics:

• Common Vertex: Both parabolas share the same vertex.
• Axes along x-axis and y-axis: One parabola has its axis along the x-axis, and the other along the y-axis.
• Intersection in the First Quadrant: This crucial detail tells us the direction in which each parabola opens. If they intersect in the first quadrant, the parabola with its axis along the x-axis must open towards the positive x-axis, and the parabola with its axis along the y-axis must open towards the positive y-axis.
• Length of Latus Rectum = 3 for both: This gives us the parameter 'a' (or 'b') for each parabola.

To solve this, we will primarily rely on:

• Standard Equations of Parabolas:
  • A parabola with vertex at $(h, k)$ and axis parallel to the x-axis has the form $(y-k)^2 = 4a(x-h)$.
  • A parabola with vertex at $(h, k)$ and axis parallel to the y-axis has the form $(x-h)^2 = 4b(y-k)$.
• Latus Rectum: The length of the latus rectum for $y^2 = 4ax$ is $|4a|$, and for $x^2 = 4by$ is $|4b|$.
• Condition for Tangency: A line $y = mx + c$ is tangent to a parabola if, when substituted into the parabola's equation, the resulting quadratic equation has a discriminant equal to zero. This signifies exactly one point of intersection.
• Slope Form of Tangents (as a shortcut/check):
  • For $y^2 = 4ax$, the tangent with slope $m$ is $y = mx + \frac{a}{m}$.
  • For $x^2 = 4by$, the tangent with slope $m$ is $y = mx - bm^2$.

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2. Step 1: Determine the Equations of the Parabolas

Concept: Identifying the vertex and using the latus rectum length to find the parameter 'a' or 'b' in the standard parabola equations.

Explanation:

1. Locate the Common Vertex: The problem states that the parabolas have a "common vertex" and their axes are along the x-axis and y-axis. The only point common to both the x-axis and y-axis is the origin $(0,0)$. Therefore, the common vertex for both parabolas is $(h,k) = (0,0)$. This simplifies our equations significantly.
2. Initial Equations based on Axes:
   • Parabola 1 (axis along x-axis, vertex at $(0,0)$): $y^2 = 4ax$.
   • Parabola 2 (axis along y-axis, vertex at $(0,0)$): $x^2 = 4by$.
3. Incorporate "First Quadrant" Information: Since the parabolas intersect in the first quadrant:
   • Parabola 1 ($y^2 = 4ax$) must open towards the positive x-axis, meaning $a > 0$.
   • Parabola 2 ($x^2 = 4by$) must open towards the positive y-axis, meaning $b > 0$.
4. Use Latus Rectum Length: We are given that the length of the latus rectum for each parabola is $3$.
   • For Parabola 1: Length of latus rectum $= |4a|$. Since $a > 0$, we have $4a = 3$, which implies $a = \frac{3}{4}$.
   • For Parabola 2: Length of latus rectum $= |4b|$. Since $b > 0$, we have $4b = 3$, which implies $b = \frac{3}{4}$.
5. Final Equations of the Parabolas:
   • Substituting $a = \frac{3}{4}$ into $y^2 = 4ax$: $y^2 = 4\left(\frac{3}{4}\right)x \implies y^2 = 3x \quad \text{(Parabola 1)}$
   • Substituting $b = \frac{3}{4}$ into $x^2 = 4by$: $x^2 = 4\left(\frac{3}{4}\right)y \implies x^2 = 3y \quad \text{(Parabola 2)}$

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3. Step 2: Set Up the General Equation of the Common Tangent

Concept: A straight line can be represented by its slope-intercept form.

Explanation: We are looking for a line that is tangent to both parabolas. Let's assume the equation of this common tangent line is $y = mx + c$, where $m$ is its slope and $c$ is its y-intercept. Our goal is to find the unique values of $m$ and $c$ that satisfy the tangency condition for both parabolas.

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4. Step 3: Apply Tangency Condition to the First Parabola ($y^2 = 3x$)

Concept: For a line to be tangent to a parabola, when their equations are combined, the resulting quadratic equation must have exactly one real root. This means its discriminant must be zero.

Explanation:

1. Substitute the Tangent Equation: We substitute the expression for $y$ from the tangent equation ($y = mx + c$) into the equation of Parabola 1 ($y^2 = 3x$). $(mx + c)^2 = 3x$
2. Form a Quadratic Equation: Expand the left side and rearrange the terms to obtain a standard quadratic equation in $x$ of the form $Ax^2 + Bx + C = 0$. $m^2x^2 + 2mcx + c^2 = 3x$ $m^2x^2 + (2mc - 3)x + c^2 = 0$
3. Apply the Discriminant Condition: For the line to be tangent, this quadratic equation must have exactly one solution for $x$. Therefore, its discriminant ($\Delta = B^2 - 4AC$) must be equal to zero. $\Delta = (2mc - 3)^2 - 4(m^2)(c^2) = 0$
4. Solve for a Relationship between $m$ and $c$: Expand and simplify the equation to find a relationship between $m$ and $c$. $4m^2c^2 - 12mc + 9 - 4m^2c^2 = 0$ $-12mc + 9 = 0$ $12mc = 9$ $c = \frac{9}{12m} \implies c = \frac{3}{4m} \quad \text{... (Equation 1)}$

Self-Check (using standard tangent form): For a parabola $y^2 = 4ax$, the tangent with slope $m$ is $y = mx + \frac{a}{m}$. Here, $a = \frac{3}{4}$. So, $c = \frac{3/4}{m} = \frac{3}{4m}$. This matches our result, confirming the algebra.

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5. Step 4: Apply Tangency Condition to the Second Parabola ($x^2 = 3y$)

Concept: The same tangency condition (discriminant equals zero) applies to the second parabola.

Explanation:

1. Substitute the Tangent Equation: Substitute $y = mx + c$ into the equation of Parabola 2 ($x^2 = 3y$). $x^2 = 3(mx + c)$
2. Form a Quadratic Equation: Expand and rearrange the terms into a quadratic equation in $x$. $x^2 = 3mx + 3c$ $x^2 - 3mx - 3c = 0$
3. Apply the Discriminant Condition: For the line to be tangent to this parabola, its discriminant must also be zero. $\Delta = (-3m)^2 - 4(1)(-3c) = 0$
4. Solve for another Relationship between $m$ and $c$: $9m^2 + 12c = 0$ $12c = -9m^2 \implies c = -\frac{9}{12}m^2 \implies c = -\frac{3}{4}m^2 \quad \text{... (Equation 2)}$

Self-Check (using standard tangent form): For a parabola $x^2 = 4by$, the tangent with slope $m$ is $y = mx - bm^2$. Here, $b = \frac{3}{4}$. So, $c = -\frac{3}{4}m^2$. This matches our result, confirming the algebra.

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6. Step 5: Solve for the Slope ($m$) and Y-intercept ($c$)

Concept: Solve the system of equations derived from the tangency conditions.

Explanation: We now have two distinct relationships between $m$ and $c$ that must both hold true for a common tangent:

1. $c = \frac{3}{4m}$
2. $c = -\frac{3}{4}m^2$

To find the values of $m$ and $c$, we can equate these two expressions for $c$: $\frac{3}{4m} = -\frac{3}{4}m^2$ To solve for $m$, we can multiply both sides by $4m$ (assuming $m \neq 0$, which is true as $c = 3/(4m)$ would be undefined otherwise). $3 = -3m^3$ $m^3 = -1$ For a real common tangent, the only real solution for $m$ is: $m = -1$

Now, substitute $m = -1$ back into either Equation 1 or Equation 2 to find $c$. Using Equation 1: $c = \frac{3}{4(-1)}$ $c = -\frac{3}{4}$

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7. Step 6: Formulate the Equation of the Common Tangent

Concept: Substitute the found values of $m$ and $c$ into the general tangent equation $y = mx + c$.

Explanation: We have determined that the slope $m = -1$ and the y-intercept $c = -\frac{3}{4}$. Substitute these values into the general equation $y = mx + c$: $y = (-1)x + \left(-\frac{3}{4}\right)$ $y = -x - \frac{3}{4}$ To match the given options, we usually clear fractions and move all terms to one side.