6. A line with a slope of 1 intersects the ellipse $x^{2}+\frac{y^{2}}{4}=1$ at points $A$ and $B$, and $P$ is a point on the line segment $AB$ such that $\frac{A P}{P B}=2$. Then the equation of the locus of point $P$ is
6. $4 x+y=\frac{2}{3} \sqrt{5-(y-x)^{2}}(|y-x|<\sqrt{5})$.

Solution: Let the moving point be $P\left(x^{\prime}, y^{\prime}\right)$, then the line through $p \quad y=x+\left(y^{\prime}-\right.$ $x^{\prime}$ ).

Substitute into the ellipse equation $x^{2}+\frac{y^{2}}{4}=1$, rearrange to get: $5 x^{2}+2\left(y^{\prime}\right.$ $\left.-x^{\prime}\right) x+\left(y^{\prime}-x^{\prime}\right)^{2}-4=0$

If the line $l$ intersects the ellipse at $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(x_{1}<\right.$ $x_{2}$ ), then $x_{1}, x_{2}$ are the two roots of equation (※), and
$$
\begin{array}{l}
x_{1}=\frac{-\left(y^{\prime}-x^{\prime}\right)-2 \sqrt{5-\left(y^{\prime}-x^{\prime}\right)^{2}}}{5} \\
x_{2}=\frac{-\left(y^{\prime}-x^{\prime}\right)+2 \sqrt{5-\left(y^{\prime}-x^{\prime}\right)^{2}}}{5} \\
\text { Also, } \because \frac{A P}{P B}=2, x_{1}<x_{2} . \therefore x^{\prime}=\frac{x_{1}+2 x_{2}}{3} .
\end{array}
$$

Substitute (1) and (2) and rearrange to get:
$$
4 x^{\prime}+y^{\prime}=\frac{2}{3} \sqrt{5-\left(y^{\prime}-x^{\prime}\right)^{2}} \quad\left(\left|y^{\prime}-x^{\prime}\right|<\right.
$$
$\sqrt{5}$ )