1. **Problem statement:** Find the area of the triangle formed by the vertex of the parabola $x^2 = 12y$ and the ends of its latus rectum. 2. **Identify the parabola parameters:** The parabola is given by $x^2 = 4ay$ where $4a = 12$, so $a = 3$. 3. **Vertex:** The vertex of the parabola is at the origin $(0,0)$. 4. **Latus rectum:** The length of the latus rectum is $4a = 12$. 5. **Coordinates of the ends of the latus rectum:** The latus rectum is a line segment perpendicular to the axis of the parabola passing through the focus. - The focus is at $(0,a) = (0,3)$. - The ends of the latus rectum are at $(rac{l}{2}, a)$ and $(-rac{l}{2}, a)$ where $l = 4a = 12$. - So the ends are at $(6,3)$ and $(-6,3)$. 6. **Vertices of the triangle:** $(0,0)$, $(6,3)$, and $(-6,3)$. 7. **Area of the triangle:** Use the formula for area with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Substitute: $$= \frac{1}{2} |0(3 - 3) + 6(3 - 0) + (-6)(0 - 3)|$$ $$= \frac{1}{2} |0 + 18 + 18| = \frac{1}{2} \times 36 = 18$$ **Final answer:** The area of the triangle is $18$ square units.