1. **Problem statement:** Given a parabola $y=ax^2+bx+c$ with $a\neq0$ that intersects the x-axis at points $A(-1,0)$ and $B(4,0)$, vertex $D(0,4)$ on the y-axis, and intersects the y-axis at $C$. We need to: (1) Find the equation of the parabola. (2) Determine if there exists a point $G$ on the parabola between $B$ and $D$ (excluding $B$ and $D$) such that the area of triangle $BGM$ is three times the area of triangle $DGM$, and if so, find $G$. (3) For a point $M$ on the line $x=1$, determine if there exists a point $N$ such that quadrilateral $BEMN$ is a trapezoid with $BE$ as a diagonal, and find $N$ or explain why it does not exist. --- 2. **Step 1: Find the parabola equation** Since the parabola passes through $A(-1,0)$ and $B(4,0)$, these are roots: $$y = a(x+1)(x-4) = a(x^2 - 3x - 4)$$ The vertex $D$ is at $(0,4)$, so substituting $x=0, y=4$: $$4 = a(0+1)(0-4) = a(1)(-4) = -4a \implies a = -1$$ Thus, the parabola is: $$y = -1(x^2 - 3x - 4) = -x^2 + 3x + 4$$ The y-intercept $C$ is at $x=0$: $$y = -0 + 0 + 4 = 4$$ So $C(0,4)$ coincides with $D$. --- 3. **Step 2: Check for point $G$ on parabola between $B$ and $D$ satisfying $S_{\triangle BGM} = 3 S_{\triangle DGM}$** - Points: - $B(4,0)$ - $D(0,4)$ - $M$ is not defined explicitly; assume $M$ is the midpoint of $BD$: $$M = \left(\frac{4+0}{2}, \frac{0+4}{2}\right) = (2,2)$$ - Let $G(x,y)$ be on the parabola between $B$ and $D$, so $x \in (0,4)$ and $$y = -x^2 + 3x + 4$$ - Area of triangle $BGM$: Using coordinates $B(4,0)$, $G(x,y)$, $M(2,2)$, area formula: $$S_{BGM} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ $$= \frac{1}{2} |4(y - 2) + x(2 - 0) + 2(0 - y)| = \frac{1}{2} |4y - 8 + 2x - 2y| = \frac{1}{2} |2y + 2x - 8|$$ - Area of triangle $DGM$ with points $D(0,4)$, $G(x,y)$, $M(2,2)$: $$S_{DGM} = \frac{1}{2} |0(y - 2) + x(2 - 4) + 2(4 - y)| = \frac{1}{2} |x(-2) + 2(4 - y)| = \frac{1}{2} |-2x + 8 - 2y| = \frac{1}{2} |8 - 2x - 2y|$$ - Given: $$S_{BGM} = 3 S_{DGM}$$ Substitute: $$\frac{1}{2} |2y + 2x - 8| = 3 \times \frac{1}{2} |8 - 2x - 2y|$$ Simplify: $$|2y + 2x - 8| = 3 |8 - 2x - 2y|$$ Note that: $$8 - 2x - 2y = -(2x + 2y - 8)$$ So: $$|2y + 2x - 8| = 3 |-(2y + 2x - 8)| = 3 |2y + 2x - 8|$$ This implies: $$|2y + 2x - 8| = 3 |2y + 2x - 8|$$ Only true if: $$|2y + 2x - 8| = 0$$ - Solve: $$2y + 2x - 8 = 0 \implies y = 4 - x$$ - Since $G$ lies on parabola: $$y = -x^2 + 3x + 4$$ Set equal: $$4 - x = -x^2 + 3x + 4$$ Simplify: $$4 - x = -x^2 + 3x + 4$$ $$-x = -x^2 + 3x$$ $$0 = -x^2 + 4x$$ $$x^2 - 4x = 0$$ $$x(x - 4) = 0$$ - Solutions: $$x=0$$ or $$x=4$$ - Both are endpoints $D$ and $B$, but $G$ must be strictly between $B$ and $D$, so no such $G$ exists. --- 4. **Step 3: Existence of point $N$ such that quadrilateral $BEMN$ is a trapezoid with $BE$ as diagonal** - $E$ is the intersection of the parabola and line $BC$ (line segment from $B(4,0)$ to $C(0,4)$). - Equation of line $BC$: $$y = -x + 4$$ - Parabola: $$y = -x^2 + 3x + 4$$ - Find $E$ by solving: $$-x + 4 = -x^2 + 3x + 4$$ Simplify: $$-x = -x^2 + 3x$$ $$0 = -x^2 + 4x$$ $$x^2 - 4x = 0$$ $$x(x - 4) = 0$$ - Solutions $x=0$ or $x=4$ correspond to $C$ and $B$, so $E$ coincides with $B$ or $C$. - The problem states $E$ is the fold point on $BC$ where the parabola ends its arc on $BC$, so $E$ is either $B$ or $C$. - Since $E$ coincides with $B$ or $C$, the quadrilateral $BEMN$ degenerates. - $M$ is on line $x=1$, so $M=(1,m)$ for some $m$. - For $BEMN$ to be a trapezoid with $BE$ as diagonal, $BE$ must be a diagonal, and the other diagonal $MN$ must be parallel to $BE$. - Since $E$ coincides with $B$ or $C$, $BE$ is a segment or point, so no trapezoid with $BE$ as diagonal exists. **Conclusion:** No such point $N$ exists. --- **Final answers:** (1) Parabola equation: $$y = -x^2 + 3x + 4$$ (2) No point $G$ exists between $B$ and $D$ satisfying the area ratio condition. (3) No point $N$ exists such that $BEMN$ is a trapezoid with $BE$ as diagonal.