1. **Problem 1: Find the standard form of the hyperbola with center (0,0), focus at (\sqrt{74},0), and directrix x=\frac{49}{\sqrt{74}}.** 2. Since the focus is on the x-axis and the directrix is vertical, the hyperbola opens horizontally. So the format is: $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ 3. The distance from the center to the focus is $c = \sqrt{74}$. 4. The directrix is at $x = \frac{49}{\sqrt{74}}$. The eccentricity $e$ is given by the ratio $e = \frac{c}{a}$. 5. The distance from the center to the directrix is $\frac{49}{\sqrt{74}}$, and by definition of eccentricity for hyperbola: $$e = \frac{c}{a} = \frac{1}{\text{distance from center to directrix}/a} = \frac{c}{a} = \frac{c}{a}$$ More precisely, the directrix distance is $\frac{a}{e}$, so: $$\frac{a}{e} = \frac{49}{\sqrt{74}}$$ 6. Using $e = \frac{c}{a}$, substitute $e$: $$\frac{a}{c/a} = \frac{49}{\sqrt{74}} \implies \frac{a^2}{c} = \frac{49}{\sqrt{74}}$$ 7. Multiply both sides by $c$: $$a^2 = c \cdot \frac{49}{\sqrt{74}} = \sqrt{74} \cdot \frac{49}{\sqrt{74}} = 49$$ 8. So, $a^2 = 49$ and $a = 7$. 9. Recall $c^2 = a^2 + b^2$ for hyperbolas, so: $$74 = 49 + b^2 \implies b^2 = 25 \implies b = 5$$ 10. **Answer for problem 1:** $$[\text{format} = \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, a=7, b=5]$$ --- 11. **Problem 2: Find vertex, focus coordinates and directrix equation for the hyperbola with center (0,0), focus (\sqrt{74},0), directrix $x=\frac{49}{\sqrt{74}}$.** 12. The vertex lies on the x-axis at distance $a$ from the center: $$ (x_c, y_c) = (0,0) $$ $$ (x_v, y_v) = (\pm a, 0) = (\pm 7, 0) $$ Since the focus is at positive $x$, take vertex at $(7,0)$. 13. Focus coordinates: $$ (x_f, y_f) = (\sqrt{74}, 0) $$ 14. Directrix equation: $$ x = \frac{49}{\sqrt{74}} $$ 15. **Answer for problem 2:** $$[x_c, y_c, x_f, y_f, eq] = [0, 0, \sqrt{74}, 0, "x = \frac{49}{\sqrt{74}}"]$$ --- 16. **Problem 3: Write equations of three circles:** - Circle A: center (1,2), radius 1 $$ (x-1)^2 + (y-2)^2 = 1^2 = 1 $$ - Circle B: center (1,0), radius 3 $$ (x-1)^2 + y^2 = 9 $$ - Circle C: center (3,0), radius 4.1 $$ (x-3)^2 + y^2 = 4.1^2 = 16.81 $$ --- 17. **Problem 4: Find asymptotes and vertices/foci of hyperbola $25y^2 - 4x^2 = 81$.** 18. Rewrite in standard form: $$ \frac{y^2}{\frac{81}{25}} - \frac{x^2}{\frac{81}{4}} = 1 $$ So: $$ a^2 = \frac{81}{25} = 3.24, \quad b^2 = \frac{81}{4} = 20.25 $$ 19. Asymptotes equations for hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ are: $$ y = \pm \frac{a}{b} x $$ Calculate: $$ \frac{a}{b} = \frac{\sqrt{3.24}}{\sqrt{20.25}} = \frac{1.8}{4.5} = 0.4 $$ So asymptotes: $$ l_1: y = 0.4 x, \quad l_2: y = -0.4 x $$ 20. Vertices are at $(0, \pm a)$: $$ V_1 = [0, 1.8], \quad V_2 = [0, -1.8] $$ 21. Foci satisfy $c^2 = a^2 + b^2$: $$ c^2 = 3.24 + 20.25 = 23.49, \quad c = \sqrt{23.49} = 4.847 $$ Foci coordinates: $$ F_1 = [0, 4.847], \quad F_2 = [0, -4.847] $$ --- **Final answers:** Problem 1: $["\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1", 7, 5]$ Problem 2: $[0, 0, \sqrt{74}, 0, "x = \frac{49}{\sqrt{74}}"]$ Problem 3: Circle A: $(x-1)^2 + (y-2)^2 = 1$ Circle B: $(x-1)^2 + y^2 = 9$ Circle C: $(x-3)^2 + y^2 = 16.81$ Problem 4: Asymptotes: $l_1: y=0.4x$, $l_2: y=-0.4x$ Vertices: $V_1=[0,1.8]$, $V_2=[0,-1.8]$ Foci: $F_1=[0,4.847]$, $F_2=[0,-4.847]$