We are given the curve $y = f(x)$ defined parametrically by the equations: $x(t) = \sin(2t) + \cos(2t), \quad y(t) = 3t^2 - 1, \quad t \in [0, 2\pi].$ We need to determine the point on the curve where the tangent is parallel to the $x$-axis. Additionally, we are tasked with finding the equations of the tangent and normal lines at that point.

Steps to Solve:

1. Find the derivative $\frac{dy}{dx}$: We first calculate the parametric derivatives: $\frac{dx}{dt} = \frac{d}{dt}[\sin(2t) + \cos(2t)] = 2\cos(2t) - 2\sin(2t),$ $\frac{dy}{dt} = \frac{d}{dt}[3t^2 - 1] = 6t.$ The slope of the tangent line is given by: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{6t}{2\cos(2t) - 2\sin(2t)} = \frac{3t}{\cos(2t) - \sin(2t)}.$ To find where the tangent is parallel to the $x$-axis, we set $\frac{dy}{dx} = 0$. This implies that: $3t = 0 \quad \Rightarrow \quad t = 0.$

2. Find the point on the curve when $t = 0$: We substitute $t = 0$ into the parametric equations for $x(t)$ and $y(t)$: $x(0) = \sin(0) + \cos(0) = 1, \quad y(0) = 3(0)^2 - 1 = -1.$ Therefore, the point on the curve is $(1, -1)$.

3. Equation of the tangent line: Since the tangent is parallel to the $x$-axis, its slope is 0. Therefore, the equation of the tangent line is: $y = -1.$

4. Equation of the normal line: The normal line is perpendicular to the tangent, which means its slope is undefined (a vertical line). Hence, the equation of the normal line is: $x = 1.$

Final Answer:

• The point where the tangent is parallel to the $x$-axis is $(1, -1)$.
• The equation of the tangent line is $y = -1$.
• The equation of the normal line is $x = 1$.

Would you like more details or have any other questions?

Here are some related questions to further explore the topic:

1. How do we interpret the geometric meaning of the normal line to a curve?
2. What if we had a different range for $t$? How would the solution change?
3. How can we find the points where the tangent is parallel to the $y$-axis?
4. How would the solution change if the parametric equations were more complex?
5. How does this approach extend to curves in three dimensions?

Tip: Always check the domain of the parameter to ensure that the points you're finding lie within the given interval.