1. **Problem Statement:** Find the coordinates of the foci, the equations of the directrices, and the length of the latus rectum for each parabola. 2. **General formulas for parabolas:** - For a parabola of the form $y^2 = 4ax$, the focus is at $(a,0)$, the directrix is $x = -a$, and the length of the latus rectum is $4a$. - For a parabola of the form $x^2 = 4ay$, the focus is at $(0,a)$, the directrix is $y = -a$, and the length of the latus rectum is $4a$. 3. **Apply to each parabola:** (i) $y^2 = 8x$ - Rewrite as $y^2 = 4(2)x$, so $a=2$. - Focus: $(2,0)$ - Directrix: $x = -2$ - Length of latus rectum: $4 \times 2 = 8$ (ii) $x^2 = 6y$ - Rewrite as $x^2 = 4(\frac{3}{2})y$, so $a=\frac{3}{2}$. - Focus: $(0, \frac{3}{2})$ - Directrix: $y = -\frac{3}{2}$ - Length of latus rectum: $4 \times \frac{3}{2} = 6$ (iii) $y^2 = -12x$ - Rewrite as $y^2 = 4(-3)x$, so $a = -3$ (parabola opens left). - Focus: $(-3,0)$ - Directrix: $x = 3$ - Length of latus rectum: $4 \times 3 = 12$ (iv) $x^2 = 16y$ - Rewrite as $x^2 = 4(4)y$, so $a=4$. - Focus: $(0,4)$ - Directrix: $y = -4$ - Length of latus rectum: $4 \times 4 = 16$ **Final answers:** (i) Focus: $(2,0)$, Directrix: $x=-2$, Latus rectum length: $8$ (ii) Focus: $(0, \frac{3}{2})$, Directrix: $y=-\frac{3}{2}$, Latus rectum length: $6$ (iii) Focus: $(-3,0)$, Directrix: $x=3$, Latus rectum length: $12$ (iv) Focus: $(0,4)$, Directrix: $y=-4$, Latus rectum length: $16$