Answer: \(\left(\frac\pi{24}-3,\frac\pi{144}-\frac12\right)\).

Answer: \(\left(\frac\pi{24}-3,\frac\pi{144}-\frac12\right)\).

Since \(P(p,-1)\) lies on the curve,

\(-1=\tan\left(3p+\frac\pi2\right).\)

Given \(0\lt p\le\frac\pi6\), we have

\(\frac\pi2\lt 3p+\frac\pi2\le\pi.\)

In this interval, \(\tan x=-1\) at \(x=\frac{3\pi}{4}\). Therefore

\(3p+\frac\pi2=\frac{3\pi}{4},\)

so

\(p=\frac\pi{12}.\)

Differentiate:

\(\frac{dy}{dx}=3\operatorname{sec}^{2}\left(3x+\frac\pi2\right).\)

At \(x=\frac\pi{12}\), the angle is \(\frac{3\pi}{4}\), so \(\operatorname{sec}^{2}\left(\frac{3\pi}{4}\right)=2\). Hence the tangent gradient is \(6\), and the normal gradient is \(-\frac16\).

The normal through \(\left(\frac\pi{12},-1\right)\) is

\(y+1=-\frac16\left(x-\frac\pi{12}\right).\)

For the \(y\)-intercept, put \(x=0\):

\(y=\frac\pi{72}-1.\)

For the \(x\)-intercept, put \(y=0\):

\(x=\frac\pi{12}-6.\)

Thus

\(A=\left(\frac\pi{12}-6,0\right),\qquad B=\left(0,\frac\pi{72}-1\right).\)

The midpoint is