1. **Problem 1: Find the standard form of the hyperbola with center (0,0), focus at (\sqrt{74},0), and directrix at x = \frac{49}{\sqrt{74}}.** 2. The hyperbola is horizontal because the focus lies on the x-axis. 3. The standard form for a horizontal hyperbola centered at the origin is: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ 4. The distance from the center to the focus is $c = \sqrt{74}$. 5. The directrix is at $x = \frac{49}{\sqrt{74}}$. 6. Recall the relationship between $a$, $b$, and $c$ for hyperbolas: $$c^2 = a^2 + b^2$$ 7. The eccentricity $e$ is defined as: $$e = \frac{c}{a}$$ 8. The directrix distance from the center is $\frac{a}{e}$, so: $$\frac{a}{e} = \frac{49}{\sqrt{74}}$$ 9. Substitute $e = \frac{c}{a}$: $$\frac{a}{c/a} = \frac{49}{\sqrt{74}} \implies \frac{a^2}{c} = \frac{49}{\sqrt{74}}$$ 10. Multiply both sides by $c$: $$a^2 = c \cdot \frac{49}{\sqrt{74}} = \sqrt{74} \cdot \frac{49}{\sqrt{74}} = 49$$ 11. So, $a^2 = 49$ and $a = 7$. 12. Calculate $b^2$ using $c^2 = a^2 + b^2$: $$74 = 49 + b^2 \implies b^2 = 25$$ 13. Therefore, the hyperbola equation is: $$\frac{x^2}{49} - \frac{y^2}{25} = 1$$ 14. The answer list is: $$["x^2/a^2 - y^2/b^2 = 1", 7, 5]$$ --- 15. **Problem 2: Find vertex, focus, and directrix of the parabola given by** $$y^2 + 10y + 25 = 64 - 16x$$ 16. Rewrite the equation: $$y^2 + 10y + 25 = 64 - 16x$$ 17. Note that $y^2 + 10y + 25 = (y+5)^2$, so: $$(y+5)^2 = 64 - 16x$$ 18. Rearrange to isolate $x$: $$16x = 64 - (y+5)^2 \implies x = 4 - \frac{(y+5)^2}{16}$$ 19. This is a parabola opening leftwards (since coefficient of $(y+5)^2$ is negative). 20. The vertex form of a horizontal parabola is: $$(y - k)^2 = 4p(x - h)$$ 21. Rewrite: $$(y+5)^2 = -16(x - 4)$$ 22. So, vertex is at $(h, k) = (4, -5)$. 23. Here, $4p = -16 \implies p = -4$. 24. Since $p$ is negative, parabola opens left. 25. Focus is at: $$(h + p, k) = (4 - 4, -5) = (0, -5)$$ 26. Directrix is the vertical line: $$x = h - p = 4 + 4 = 8$$ 27. The answer list is: $$[4, -5, 0, -5, "x=8"]$$ --- 28. **Problem 3: Write equations of circles A, B, and C given centers and approximate radii.** 29. Circle A: center at $(1, 2)$, small radius. Assume radius $r_A$. 30. Circle B: center at $(1, 0)$, larger radius $r_B$. 31. Circle C: center at $(3, -2)$, largest radius $r_C$. 32. Without exact radii, we cannot give precise equations, but general forms are: $$\text{Circle A}: (x - 1)^2 + (y - 2)^2 = r_A^2$$ $$\text{Circle B}: (x - 1)^2 + (y - 0)^2 = r_B^2$$ $$\text{Circle C}: (x - 3)^2 + (y + 2)^2 = r_C^2$$ --- 33. **Problem 4: For hyperbola $25y^2 - 4x^2 = 81$, find asymptotes, vertices, and foci.** 34. Rewrite in standard form: $$\frac{y^2}{\frac{81}{25}} - \frac{x^2}{\frac{81}{4}} = 1$$ 35. So, $$a^2 = \frac{81}{25} = 3.24, \quad b^2 = \frac{81}{4} = 20.25$$ 36. The hyperbola opens vertically because $y^2$ term is positive. 37. Asymptotes for vertical hyperbola: $$y = \pm \frac{a}{b} x$$ 38. Calculate: $$\frac{a}{b} = \frac{\sqrt{3.24}}{\sqrt{20.25}} = \frac{1.8}{4.5} = 0.4$$ 39. Asymptotes: $$l_1: y = 0.4x, \quad l_2: y = -0.4x$$ 40. Vertices are at $(0, \pm a)$: $$V_1 = [0, 1.8], \quad V_2 = [0, -1.8]$$ 41. Calculate $c$ for foci: $$c^2 = a^2 + b^2 = 3.24 + 20.25 = 23.49$$ $$c = \sqrt{23.49} = 4.847$$ 42. Foci at $(0, \pm c)$: $$F_1 = [0, 4.847], \quad F_2 = [0, -4.847]$$ --- **Final answers:** 1. Hyperbola standard form, $a$, $b$: $$["x^2/a^2 - y^2/b^2 = 1", 7, 5]$$ 2. Parabola vertex, focus, directrix: $$[4, -5, 0, -5, "x=8"]$$ 3. Circles equations (general form): $$[(x-1)^2 + (y-2)^2 = r_A^2, (x-1)^2 + y^2 = r_B^2, (x-3)^2 + (y+2)^2 = r_C^2]$$ 4. Hyperbola asymptotes, vertices, foci: $$l_1: y=0.4x, l_2: y=-0.4x$$ $$V_1=[0,1.8], V_2=[0,-1.8]$$ $$F_1=[0,4.847], F_2=[0,-4.847]$$