Example 3 Let the curve $C_{1}: \frac{x^{2}}{3}+\frac{y^{2}}{4}=1$, curve $C_{2}: x+y=1$, and $P$ be a point on $C_{2}$. The ray $O P$ intersects $C_{1}$ at $R$. Also, point $Q$ is on $O P$, and satisfies $|O Q| \cdot|O P|=|O R|^{2}$. Find the equation of the trajectory of point $Q$ as $P$ moves along the curve $C_{2}$.
Let $f(x, y)=\frac{x^{2}}{3}+\frac{y^{2}}{4}, g(x, y)=x+y$.
Notice property 9, and $f(t x, t y)=\frac{(t x)^{2}}{3}+\frac{(t y)^{2}}{4}=t^{2}\left(\frac{x^{2}}{3}+\frac{y^{2}}{4}\right)=t^{2} f(x, y), g(t x, t y)=$ $t x+t y=t \cdot g(x, y)$, thus the trajectory equation of point $Q$ is $\frac{x^{2}}{3}+\frac{y^{2}}{4}=x+y(x+y>0)$, which simplifies to $\frac{\left(x-\frac{3}{2}\right)^{2}}{\frac{21}{4}}+\frac{(y-2)^{2}}{7}=1$ (excluding the origin).