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To solve this step-by-step, we are tasked with finding the equation of the normal to the curve \( x = 2 \tan\left(y + \frac{\pi}{12}\right) \) at the point \( P \) with \( y \)-coordinate \( \frac{\pi}{4} \).

---

### Step 1: Differentiate \( x \) with respect to \( y \)
The given equation is:
\[
x = 2 \tan\left(y + \frac{\pi}{12}\right).
\]

Differentiating both sides with respect to \( y \):
\[
\frac{dx}{dy} = 2 \cdot \sec^2\left(y + \frac{\pi}{12}\right).
\]

This gives the derivative of \( x \) with respect to \( y \), which is:
\[
\frac{dx}{dy} = 2 \sec^2\left(y + \frac{\pi}{12}\right).
\]

---

### Step 2: Evaluate the derivative at \( y = \frac{\pi}{4} \)
At \( y = \frac{\pi}{4} \), substitute into the derivative:
\[
\frac{dx}{dy} = 2 \sec^2\left(\frac{\pi}{4} + \frac{\pi}{12}\right).
\]

Simplify \( \frac{\pi}{4} + \frac{\pi}{12} \) to a common denominator:
\[
\frac{\pi}{4} + \frac{\pi}{12} = \frac{3\pi}{12} + \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}.
\]

So:
\[
\frac{dx}{dy} = 2 \sec^2\left(\frac{\pi}{3}\right).
\]

From trigonometric identities, \( \sec\left(\frac{\pi}{3}\right) = 2 \), so:
\[
\frac{dx}{dy} = 2 (2^2) = 2 \cdot 4 = 8.
\]

Thus:
\[
\frac{dx}{dy} = 8.
\]

---

### Step 3: Find the slope of the normal
The slope of the curve at the point \( P \) is \( \frac{dx}{dy} = 8 \). The slope of the normal is the negative reciprocal of this:
\[
\text{slope of normal} = -\frac{1}{8}.
\]

---

### Step 4: Find the coordinates of point \( P \)
We know that \( y = \frac{\pi}{4} \). Substitute \( y = \frac{\pi}{4} \) into the original equation \( x = 2 \tan\left(y + \frac{\pi}{12}\right) \):
\[
x = 2 \tan\left(\frac{\pi}{4} + \frac{\pi}{12}\right).
\]

We already calculated \( \frac{\pi}{4} + \frac{\pi}{12} = \frac{\pi}{3} \), so:
\[
x = 2 \tan\left(\frac{\pi}{3}\right).
\]

From trigonometric identities, \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \), so:
\[
x = 2 \sqrt{3}.
\]

Thus, the coordinates of \( P \) are:
\[
P = \left(2\sqrt{3}, \frac{\pi}{4}\right).
\]

---

### Step 5: Equation of the normal
The equation of a straight line is:
\[
y - y_1 = m(x - x_1),
\]
where \( m \) is the slope, and \( (x_1, y_1) \) is the point.

Here:
\[
m = -\frac{1}{8}, \quad (x_1, y_1) = \left(2\sqrt{3}, \frac{\pi}{4}\right).
\]

Substitute:
\[
y - \frac{\pi}{4} = -\frac{1}{8}(x - 2\sqrt{3}).
\]

Simplify:
\[
y = -\frac{1}{8}x + \frac{2\sqrt{3}}{8} + \frac{\pi}{4}.
\]

Further simplify:
\[
y = -\frac{1}{8}x + \frac{\sqrt{3}}{4} + \frac{\pi}{4}.
\]

This is the equation of the normal.

---

### Final Answer:
The equation of the normal to the curve at \( P \) is:
\[
y = -\frac{1}{8}x + \frac{\sqrt{3} + \pi}{4}.
\]

---

Would you like a deeper explanation or have additional questions?

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### Follow-Up Questions:
1. How is the reciprocal of the slope used in the normal calculation?
2. Can you show the trigonometric simplifications for \( \sec(\pi/3) \) and \( \tan(\pi/3) \)?
3. How do we handle implicit differentiation for curves not expressed as \( y = f(x) \)?
4. What are the key differences between the tangent and the normal at a point on a curve?
5. Can this method be generalized for other trigonometric curves?

---

### Tip:
Always keep track of intermediate trigonometric results, as errors can propagate through simplifications.

The point P is the point on the curve x = 2tan(y + π/12) with y-coordinate π/4. Find an equation of the normal to the curve at P.

Learn how to find the equation of the normal to the curve x = 2tan(y + π/12) at the point with y-coordinate π/4. This step-by-step solution involves differentiation, evaluating trigonometric functions, and deriving the equation of the normal line.

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