帮我做一下这个题的讲解这是关于抛物线跟椭圆的---**Problem Information:** * Problem Number: 30. * Points: (本题 10 分) - 10 points **Question Stem:** 已知抛物线的顶点在坐标原点，椭圆 $\frac{x^2}{4} + y^2 = 1$ 的顶点分别为 $A_1, A_2, B_1, B_2$，其中点 $A_2$ 为抛物线的焦点，如图所示。 **Sub-questions:** (1) 求抛物线的标准方程； (2) 若过点 $A_1$ 的直线 $l$ 与抛物线交于 $M, N$ 两点，且 $(\vec{OM} + \vec{ON}) // \vec{B_1A_2}$，求直线 $l$ 的 **Diagram Description:** * **Type:** Coordinate plane with an ellipse and a parabola. * **Coordinate Axes:** Standard x-axis and y-axis intersecting at the origin O. The x-axis is horizontal pointing right, the y-axis is vertical pointing up. * **Origin:** Labeled as O at the intersection of the axes. * **Ellipse:** Centered at the origin O. It is wider than it is tall. * Vertices are labeled $A_1$, $A_2$, $B_1$, $B_2$. * $A_1$ is on the negative x-axis. * $A_2$ is on the positive x-axis. * $B_2$ is on the positive y-axis. * $B_1$ is on the negative y-axis. * The equation of the ellipse is given as $\frac{x^2}{4} + y^2 = 1$. From this, the semi-major axis is $a=2$ (along the x-axis) and the semi-minor axis is $b=1$ (along the y-axis). * Coordinates of vertices: $A_1(-2, 0)$, $A_2(2, 0)$, $B_1(0, -1)$, $B_2(0, 1)$. * **Parabola:** Vertex at the origin O. * The diagram shows a parabola opening to the right. * The text states that $A_2$ is the focus of the parabola. * A handwritten equation $y^2=4x$ is written next to the diagram, which is consistent with a parabola with vertex at (0,0) and focus at (1,0), or (2,0) if $p=2$. Given that $A_2$ is the focus and $A_2=(2,0)$, the focus is at $(2,0)$. The vertex is at $(0,0)$. Therefore, the focal length $p=2$. The standard equation for a parabola opening to the right with vertex at the origin is $y^2 = 4px$. So, the equation is $y^2 = 4(2)x = 8x$. *Note: The handwritten equation $y^2=4x$ next to the diagram might be a different parabola or incorrect for the problem statement where $A_2(2,0)$ is the focus.* We should rely on the text description. * **Other Elements:** * A line $l$ passing through $A_1$ is shown intersecting the parabola at two points, labeled M and N. * Vectors $\vec{OM}$ and $\vec{ON}$ are implied by the condition $(\vec{OM} + \vec{ON}) // \vec{B_1A_2}$. * The vector $\vec{B_1A_2}$ is implied by the points $B_1(0,-1)$ and $A_2(2,0)$. $\vec{B_1A_2} = A_2 - B_1 = (2,0) - (0,-1) = (2, 1)$. **Handwritten Annotation:** * Next to the diagram, handwritten text "y^2=4x" is present. * Above the problem number, handwritten text "2√3" is present.