13.6.9 ** Find two points \( P \) and \( Q \) on the curve \( y^2 = x \) such that the tangent line to the curve at point \( P \) is perpendicular to the line \( PQ \), and the tangent line to the curve at point \( Q \) forms an angle of \( \frac{\pi}{4} \) with the line \( PQ \) (draw a sketch).
Let the coordinates of the points be $P\left(a^{2}, a\right)$ and $Q\left(b^{2}, b\right)$. The slope of the tangent line $l_{1}$ to the curve $y^{2}=x$ at point $P$ is $k_{1}=\frac{1}{2} a$, and the slope of the tangent line $l_{2}$ at point $Q$ is $k_{2}=\frac{1}{2} b$.

Since $P Q \parallel l_{1}$, the slope of $P Q$ is $k=-2 a$. Given that the angle between line $l_{2}$ and $P Q$ is $\frac{\pi}{4}$, we have
$$
\tan \frac{\pi}{4}= \pm \frac{k_{2}-k}{1+k k_{2}}= \pm \frac{\frac{1}{2 b}+2 a}{1-2 a \cdot \frac{1}{2 b}},
$$

which simplifies to
$$
4 a b-2 a+2 b+1=0
$$

or
$$
4 a b+2 a-2 b+1=0,
$$
(2)
(3)
Since $P Q$ passes through $P\left(a^{2}, a\right)$ and $Q\left(b^{2}, b\right)$, and the slope $k=-2 a$, we have
$$
\frac{b-a}{b^{2}-a^{2}}=-2 a \text {, which simplifies to } 2 a^{2}+2 a b+1=0 .
$$

Combining (3) with (1) and (2) respectively, we find the points $P$ and $Q$ that satisfy the conditions to be
$$
(1,-1),\left(\frac{9}{4}, \frac{3}{2}\right) \text { or }(1,1),\left(\frac{9}{4},-\frac{3}{2}\right) \text {. }
$$