1. **Stating the problem:** We are given an ellipse defined by the equation $$\frac{x^2}{6} + \frac{y^2}{12} = 1$$ and two points $P(2, 2)$ and $Q\left(\sqrt{\frac{3}{2}}, 3\right)$. 2. **Goal:** Determine the position of points $P$ and $Q$ relative to the ellipse (inside, on, or outside). 3. **Formula and rule:** For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, a point $(x_0, y_0)$ is: - Inside if $$\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} < 1$$ - On the ellipse if $$\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1$$ - Outside if $$\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} > 1$$ Here, $a^2 = 6$ and $b^2 = 12$. 4. **Check point P(2, 2):** Calculate $$\frac{2^2}{6} + \frac{2^2}{12} = \frac{4}{6} + \frac{4}{12} = \frac{4}{6} + \frac{1}{3} = \frac{4}{6} + \frac{2}{6} = \frac{6}{6} = 1$$ Since the sum equals 1, point $P$ lies exactly on the ellipse. 5. **Check point Q\left(\sqrt{\frac{3}{2}}, 3\right):** Calculate $$\frac{\left(\sqrt{\frac{3}{2}}\right)^2}{6} + \frac{3^2}{12} = \frac{\frac{3}{2}}{6} + \frac{9}{12} = \frac{3}{12} + \frac{9}{12} = \frac{12}{12} = 1$$ Since the sum equals 1, point $Q$ also lies exactly on the ellipse. **Final answer:** Both points $P(2, 2)$ and $Q\left(\sqrt{\frac{3}{2}}, 3\right)$ lie on the ellipse defined by $$\frac{x^2}{6} + \frac{y^2}{12} = 1$$.