We are tasked with finding the parametric equations of the curve formed by the intersection of the cone $z = \sqrt{x^2 + y^2}$ and the plane $z = 4 + y$.

Step 1: Set up the equations

The cone equation is given as: $z = \sqrt{x^2 + y^2}$ The plane equation is: $z = 4 + y$

Step 2: Equating the two surfaces

At the points of intersection, the $z$-values from both surfaces must be equal. So, we equate the two equations: $\sqrt{x^2 + y^2} = 4 + y$ To eliminate the square root, square both sides: $x^2 + y^2 = (4 + y)^2$ Expanding the right-hand side: $x^2 + y^2 = 16 + 8y + y^2$ Canceling $y^2$ from both sides: $x^2 = 16 + 8y$ Thus, we now have the relationship: $x^2 = 16 + 8y$

Step 3: Parametrize $y$

We can parametrize $y$ with a parameter $t$. A simple choice is: $y(t) = t$ Thus, we substitute $y = t$ into the equation for $x^2$: $x^2 = 16 + 8t$ This gives us: $x(t) = \pm\sqrt{16 + 8t}$

For simplicity, let's choose $x(t) = \sqrt{16 + 8t}$ as one branch of the curve.

Step 4: Parametrize $z$

From the plane equation, $z = 4 + y$, we know that: $z(t) = 4 + t$

Final Parametric Equations

The parametric equations of the curve are: $x(t) = \sqrt{16 + 8t}$ $y(t) = t$ $z(t) = 4 + t$

These describe the curve of intersection between the cone and the plane.

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Would you like more details on a specific step or have any further questions?

Here are five related questions to explore:

1. How would the parametric equations change if the plane equation were different?
2. What is the geometric interpretation of the curve in 3D space?
3. Can the parametric equations be written in terms of trigonometric functions?
4. How does the curve behave as $t \to \infty$?
5. What would happen if the cone had a different orientation?

Tip: Always check for extraneous solutions when squaring both sides of an equation.