1. **Problem Statement:** Given the parabola equation $y^2 = 8x$, find the coordinates of the focus, the endpoints of the latus rectum, and the equation of the directrix. Then sketch the parabola. 2. **Formula and Important Rules:** The standard form of a parabola that opens rightward is $y^2 = 4ax$, where $a$ is the distance from the vertex to the focus. - The vertex is at the origin $(0,0)$. - The focus is at $(a,0)$. - The directrix is the vertical line $x = -a$. - The length of the latus rectum is $4a$. - The endpoints of the latus rectum lie on the line through the focus perpendicular to the axis of symmetry. 3. **Identify $a$ from the equation:** Given $y^2 = 8x$, compare with $y^2 = 4ax$. So, $4a = 8$ which gives $a = 2$. 4. **Find the focus:** Focus is at $(a,0) = (2,0)$. 5. **Find the directrix:** Directrix is the line $x = -a = -2$. 6. **Find the endpoints of the latus rectum:** Length of latus rectum = $4a = 8$. Since the latus rectum is horizontal at the focus, endpoints are at: $(2, 4)$ and $(2, -4)$ if vertical axis, but here axis is horizontal, so endpoints are at $(2, 2)$ and $(2, -2)$ because the parabola opens rightward and $y^2=8x$ means vertical distance from focus is $2a=4$, but since $4a=8$, half length is $4$, so endpoints are $(2, 4)$ and $(2, -4)$. However, the problem states endpoints at $(2, 2)$ and $(2, -2)$, so let's verify: The latus rectum length is $4a=8$, so half length is $4$, so endpoints should be $(2, 4)$ and $(2, -4)$. But the problem's graph shows $(2, 2)$ and $(2, -2)$, so likely the problem uses $a=1$ or a different scale. Given the equation $y^2=8x$, $a=2$, so endpoints are $(2, 4)$ and $(2, -4)$. 7. **Summary:** - Focus: $(2,0)$ - Directrix: $x = -2$ - Endpoints of latus rectum: $(2,4)$ and $(2,-4)$ 8. **Sketch:** The parabola opens to the right with vertex at origin, focus at $(2,0)$, directrix at $x=-2$, and latus rectum endpoints at $(2,4)$ and $(2,-4)$. Final answers: Focus: $(2,0)$ Directrix: $x = -2$ Endpoints of latus rectum: $(2,4)$ and $(2,-4)$