1. The problem is to analyze the parabola given by the equation $$-12x - 36 = (y - 6)^2$$ and verify the focus, vertex, and endpoints of the latus rectum. 2. First, rewrite the equation in the standard form of a parabola. Start by isolating $x$: $$-12x - 36 = (y - 6)^2$$ $$-12x = (y - 6)^2 + 36$$ $$x = -\frac{(y - 6)^2 + 36}{12}$$ 3. Simplify the expression: $$x = -\frac{(y - 6)^2}{12} - \frac{36}{12}$$ $$x = -\frac{(y - 6)^2}{12} - 3$$ 4. This matches the form of a horizontal parabola: $$ (y - k)^2 = 4p(x - h) $$ where the vertex is at $(h, k)$. 5. Comparing, we have: $$ (y - 6)^2 = -12(x + 3) $$ So, $$ h = -3, \quad k = 6, \quad 4p = -12 \Rightarrow p = -3 $$ 6. The vertex is at $(-3, 6)$, which matches the given vertex. 7. Since $p = -3$, the parabola opens to the left (negative $x$ direction). 8. The focus is located at: $$ (h + p, k) = (-3 - 3, 6) = (-6, 6) $$ which matches the given focus. 9. The latus rectum length is $|4p| = 12$. The endpoints of the latus rectum are located $\frac{|4p|}{2} = 6$ units above and below the focus along the $y$-axis: $$ (-6, 6 + 6) = (-6, 12) $$ $$ (-6, 6 - 6) = (-6, 0) $$ However, the problem states the endpoints as $(-6, 9)$ and $(-6, 3)$, which are 3 units above and below the focus. 10. This suggests a discrepancy; the correct endpoints of the latus rectum should be $(-6, 12)$ and $(-6, 0)$. Final answer: - Vertex: $(-3, 6)$ - Focus: $(-6, 6)$ - Endpoints of latus rectum: $(-6, 12)$ and $(-6, 0)$ The given points $(-6, 9)$ and $(-6, 3)$ are not the correct endpoints of the latus rectum for this parabola.