To solve the problem:

Step 1: Understand the problem

The equation of the curve is: $x = 2 \tan\left(y + \frac{\pi}{12}\right)$ We need to find the equation of the normal to the curve at the point $P$, where the $y$-coordinate is $\frac{\pi}{4}$.

Step 2: Calculate $x$-coordinate for $y = \frac{\pi}{4}$

Substitute $y = \frac{\pi}{4}$ into the curve equation: $x = 2 \tan\left(\frac{\pi}{4} + \frac{\pi}{12}\right)$ Simplify: $\frac{\pi}{4} + \frac{\pi}{12} = \frac{3\pi}{12} + \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$ Thus: $x = 2 \tan\left(\frac{\pi}{3}\right)$ The value of $\tan\left(\frac{\pi}{3}\right)$ is $\sqrt{3}$, so: $x = 2\sqrt{3}$ Therefore, the point $P$ is $\left(2\sqrt{3}, \frac{\pi}{4}\right)$.

Step 3: Differentiate the curve to find the gradient

The curve is given by: $x = 2 \tan\left(y + \frac{\pi}{12}\right)$ Differentiate both sides with respect to $y$: $\frac{dx}{dy} = 2 \sec^2\left(y + \frac{\pi}{12}\right)$ The gradient of the tangent line is: $\frac{dx}{dy} = 2 \sec^2\left(\frac{\pi}{4} + \frac{\pi}{12}\right)$ Substitute $y = \frac{\pi}{4}$, which we already know gives $y + \frac{\pi}{12} = \frac{\pi}{3}$. Thus: $\frac{dx}{dy} = 2 \sec^2\left(\frac{\pi}{3}\right)$ The value of $\sec\left(\frac{\pi}{3}\right)$ is $2$, so: $\frac{dx}{dy} = 2 (2^2) = 8$

Step 4: Gradient of the normal

The gradient of the normal is the negative reciprocal of the tangent's gradient: $\text{Gradient of normal} = -\frac{1}{8}$

Step 5: Equation of the normal

The equation of a straight line is: $y - y_1 = m(x - x_1)$ Here, $m = -\frac{1}{8}$, $(x_1, y_1) = (2\sqrt{3}, \frac{\pi}{4})$. Substituting these values: $y - \frac{\pi}{4} = -\frac{1}{8}(x - 2\sqrt{3})$ Simplify: $y = -\frac{1}{8}x + \frac{\sqrt{3}}{4} + \frac{\pi}{4}$

This is the equation of the normal: $y = -\frac{1}{8}x + \frac{\sqrt{3} + \pi}{4}$

Let me know if you'd like further clarifications or details.
Here are some related questions to deepen your understanding:

1. What is the physical meaning of the gradient of a curve?
2. How do you find the equation of a tangent for parametric equations?
3. What is the role of reciprocal gradients in normal lines?
4. Why is the point $P$ important in solving this problem?
5. How would the solution differ if the curve were implicit?

Tip: Always verify trigonometric values for standard angles like $\frac{\pi}{3}$ to ensure accuracy in calculations.