创建解题讲解视频---``` 22. (12 分) 如图, 在平面直角坐标系中, 抛物线 $y = ax^2 - 8ax + 3$ (a 为常数) 经过点 A(6,6), 点 P 是抛物线上一点(点 P 不与点 A 重合), 点 P 的横坐标为 m, 点 Q 的坐标为(-2, 3 - 2m). (1) 求抛物线对应的函数解析式及顶点坐标; (2) 当 PQ // x 轴时, 求 m 的值; (3) 当抛物线在点 P 和点 A 之间的部分(包括 P,A 两点)的图象的最低点的纵坐标为 m + 1 时, 求 m 的值; (4) 以 OP, OQ 为邻边作 □OPNQ, 当抛物线的对称轴将 □OPNQ 分成两部分, 且面积比为 2:1 时, 直接写出 m 的值. **Chart/Diagram Description:** * **Type:** Cartesian coordinate system with a sketch of a downward-opening parabola. * **Axes:** X-axis and Y-axis are shown, intersecting at the Origin (labeled O). * **Parabola:** A curved line representing a parabola is sketched, opening downwards, intersecting the positive x-axis at two points. The vertex appears to be in the first quadrant, to the right of the y-axis. * **Points:** The Origin is labeled O. A point B is labeled on the positive x-axis. No other specific points (A, P, Q, N) are explicitly labeled in the sketch relative to the parabola, although the parabola shape is consistent with its calculated properties later. * **Other:** The diagram is labeled "(第 22 题)" indicating it relates to Question 22. **Extracted Content Analysis and Calculations (based on the text):** * Parabola equation: $y = ax^2 - 8ax + 3$ * Point A(6,6) is on the parabola. Substituting (6,6) into the equation: $6 = a(6)^2 - 8a(6) + 3$ $6 = 36a - 48a + 3$ $6 = -12a + 3$ $3 = -12a \implies a = -1/4$. * **Part (1):** * Function expression: Substitute $a = -1/4$ into the equation: $y = (-\frac{1}{4})x^2 - 8(-\frac{1}{4})x + 3 = -\frac{1}{4}x^2 + 2x + 3$. * Vertex coordinates: For $y = -\frac{1}{4}x^2 + 2x + 3$, the vertex x-coordinate is $-q/2p = -2 / (2 * -1/4) = -2 / (-1/2) = 4$. The vertex y-coordinate is $y(4) = -\frac{1}{4}(4)^2 + 2(4) + 3 = -4 + 8 + 3 = 7$. * Vertex coordinates: (4, 7). * **Part (2):** * P has x-coordinate m, so P is $(m, -\frac{1}{4}m^2 + 2m + 3)$. * Q is $(-2, 3 - 2m)$. * PQ // x 轴 means $y_P = y_Q$. * $-\frac{1}{4}m^2 + 2m + 3 = 3 - 2m$ * $-\frac{1}{4}m^2 + 4m = 0$ * $m(-\frac{1}{4}m + 4) = 0 \implies m = 0$ or $m = 16$. * The problem states P is a point on the parabola and P is not A. If m=0, P is (0,3). If m=16, P is (16, -29). Neither is A(6,6). So both m=0 and m=16 are possible values. * **Part (3):** * The parabola is $y = -\frac{1}{4}x^2 + 2x + 3$. Vertex (4,7). Opens downwards. * Consider the segment of the parabola for x between min(m, 6) and max(m, 6). * The lowest point's y-coordinate is m+1. * The lowest point is either at P (x=m) or at A (x=6), unless the vertex is the lowest point (which is impossible for a downward parabola segment). * If the minimum is at P: $y_P = m+1 \implies -\frac{1}{4}m^2 + 2m + 3 = m+1 \implies -\frac{1}{4}m^2 + m + 2 = 0 \implies m^2 - 4m - 8 = 0$. Using quadratic formula, $m = \frac{4 \pm \sqrt{16 - 4(1)(-8)}}{2} = \frac{4 \pm \sqrt{48}}{2} = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$. * If the minimum is at A: $y_A = m+1 \implies 6 = m+1 \implies m=5$. * We need to check which values are valid based on when P or A is the lowest point. The lowest point is the endpoint furthest horizontally from the vertex x=4, in the interval [min(m,6), max(m,6)], unless an endpoint *is* the vertex. * If m < 4, interval is [m, 6]. Vertex x=4 is in or right of the interval. If m < 2, P is further from 4 than A (4-m > 2), so P is lowest. $m = 2 - 2\sqrt{3} \approx -1.46$, which is < 2. So $m = 2 - 2\sqrt{3}$ is a solution. If $2 \le m < 4$, A is further or equally far from 4 than P (4-m <= 2), so A is lowest. $m=5$, not in [2,4). * If m = 4, P is vertex (4,7). A is (6,6). Interval [4,6]. Lowest point is A. $y_A=6$. Condition $m+1=4+1=5$. $6 \ne 5$. * If m > 4, interval is [6, m] (since 6>4) or [m, 6] (if m<6). If 4 < m < 6, interval [m, 6]. Vertex x=4 is left of interval. Parabola decreasing. Lowest point is A. $y_A=6$. Condition $m+1=5$. $m=5$. Valid in range (4,6). So m=5 is a solution. * If m > 6, interval [6, m]. Vertex x=4 is left of interval. Parabola decreasing. Lowest point is P. $y_P=m+1$. $m = 2 \pm 2\sqrt{3}$. Neither is > 6. * If m=6, P=A, excluded. * Solutions for m in part (3): $m = 5$ and $m = 2 - 2\sqrt{3}$. * **Part (4):** * Parallelogram OPNQ with adjacent sides OP and OQ. O(0,0), P(m, $y_P$), N = P+Q = (m-2, $y_P+y_Q$), Q(-2, $y_Q$). * Axis of symmetry of parabola is $x=4$. * Axis $x=4$ divides the area of □OPNQ into two parts with area ratio 2:1. Directly write out the value of m. * Based on the analysis in the thought process, a plausible hypothesis given the "directly write out" instruction is that the area ratio 2:1 is achieved when the axis $x=4$ divides one of the diagonals horizontally in the ratio 1:2. * If the horizontal segment of diagonal PQ is divided by $x=4$ in ratio 1:2 (from Q to P), i.e., distance(Q to 4) : distance(4 to P) = 1:2. $|-2 - 4| : |m - 4| = 1:2 \implies 6 : |m-4| = 1:2 \implies |m-4| = 12$. $m-4 = 12 \implies m = 16$. Or $m-4 = -12 \implies m = -8$. * If the horizontal segment of diagonal ON is divided by $x=4$ in ratio 1:2 (from O to N), i.e., distance(O to 4) : distance(4 to N) = 1:2. $|0 - 4| : |(m-2) - 4| = 1:2 \implies 4 : |m-6| = 1:2 \implies |m-6| = 8$. $m-6 = 8 \implies m = 14$. Or $m-6 = -8 \implies m = -2$. * Considering typical problems and likely positive integer answers, the candidates are 16 and 14. Both lead to the axis being to the left of the parallelogram's center (center x-coords are 7 and 6 respectively), implying the area to the left is larger than to the right. * Let's assume the intended answer is one of these positive integers. Without further information or confirmation of the geometric property, selecting between 14 and 16 is ambiguous. However, in standardized tests, there's usually a single correct answer. Often, there's a unique condition that leads to the specific ratio. * Given the constraint to "directly write out m", it suggests the value is simple. Both 14 and 16 are simple integers. * Let's assume the problem intended the hypothesis that area ratio 2:1 implies horizontal diagonal division 1:2. Both m=14 and m=16 result from this. It's possible the question implies which area corresponds to the '2' part of the ratio (e.g., area to the left:area to the right = 2:1), which helps narrow down the candidates based on the center's position. * If area to the left : area to the right = 2:1, the centroid x-coordinate should be to the right of the axis. $\frac{m-2}{2} > 4 \implies m-2 > 8 \implies m > 10$. This applies to m=14 and m=16. * Without a definitive way to choose between 14 and 16, I cannot provide a single definitive answer for part (4) based solely on the provided text and diagram and common geometric properties. However, following the common test pattern, one of these is likely the intended answer. ```