**A. Given the equations, find the following:** **Problem 1: $y^2 - 25x^2 + 4y + 50x - 46 = 0$** 1. Rewrite the equation grouping $x$ and $y$ terms: $$y^2 + 4y - 25x^2 + 50x = 46$$ 2. Complete the square for $y$: $$y^2 + 4y = (y+2)^2 - 4$$ 3. Complete the square for $x$: $$-25x^2 + 50x = -25(x^2 - 2x) = -25[(x-1)^2 -1] = -25(x-1)^2 + 25$$ 4. Substitute completed squares: $$(y+2)^2 - 4 - 25(x-1)^2 + 25 = 46$$ 5. Simplify constants: $$(y+2)^2 - 25(x-1)^2 + 21 = 46 ightarrow (y+2)^2 - 25(x-1)^2 = 25$$ 6. Divide both sides by 25: $$\frac{(y+2)^2}{25} - \frac{(x-1)^2}{1} = 1$$ This is the standard form of a vertical hyperbola. **a. Standard equation:** $$\frac{(y+2)^2}{25} - \frac{(x-1)^2}{1} = 1$$ **b. Vertices:** Center: $(1, -2)$ Vertices are $a=5$ units from center along $y$-axis: $$(1, -2 \pm 5) = (1, 3) \text{ and } (1, -7)$$ **c. Co-vertices:** Along $x$-axis, distance $b=1$ from center: $$(1 \pm 1, -2) = (0, -2) \text{ and } (2, -2)$$ **d. Foci:** $$c^2 = a^2 + b^2 = 25 + 1 = 26$$ $$c=\sqrt{26}$$ Foci at $(1, -2 \pm \sqrt{26})$ **e. Length of axes:** Major axis: $2a = 10$ Minor axis: $2b = 2$ **f. Asymptotes:** For vertical hyperbola, $$y - k = \pm \frac{a}{b} (x - h)$$ $$y + 2 = \pm 5 (x - 1)$$ **g. Eccentricity:** $$e = \frac{c}{a} = \frac{\sqrt{26}}{5}$$ --- **Problem 2: $\frac{x^2}{144} - \frac{y^2}{49} = 1$** (horizontal hyperbola) **a. Standard equation:** Given **b. Vertices:** Center at origin $(0,0)$ $a=12$, vertices at $(\pm 12, 0)$ **c. Co-vertices:** $b=7$, co-vertices at $(0, \pm 7)$ **d. Foci:** $$c^2 = a^2 + b^2 = 144 + 49 = 193$$ $$c = \sqrt{193}$$ Foci at $(\pm \sqrt{193}, 0)$ **e. Length of axes:** Major axis: $2a = 24$ Minor axis: $2b = 14$ **f. Asymptotes:** $$y = \pm \frac{b}{a} x = \pm \frac{7}{12}x$$ **g. Eccentricity:** $$e = \frac{c}{a} = \frac{\sqrt{193}}{12}$$ --- **Problem 3: $\frac{x^2}{9} + \frac{y^2}{25} = 1$** (ellipse) **a. Standard equation:** Given **b. Vertices:** Center at origin $(0,0)$ $a=5$ (major axis along $y$), vertices at $(0, \pm 5)$ **c. Co-vertices:** $b=3$, co-vertices at $(\pm 3, 0)$ **d. Foci:** $$c^2 = a^2 - b^2 = 25 - 9 = 16$$ $$c=4$$ Foci at $(0, \pm 4)$ **e. Length of axes:** Major axis: $2a=10$ Minor axis: $2b=6$ **f. Asymptotes:** Not applicable to ellipse **g. Eccentricity:** $$e = \frac{c}{a} = \frac{4}{5}$$ --- **Problem 4: $x^2 - 2x + 4y^2 - 16y = -13$** (ellipse) 1. Group terms: $$x^2 - 2x + 4(y^2 - 4y) = -13$$ 2. Complete the square for $x$: $$x^2 - 2x = (x-1)^2 - 1$$ 3. Complete the square for $y$: $$y^2 - 4y = (y-2)^2 - 4$$ 4. Substitute and expand: $$(x-1)^2 -1 + 4[(y-2)^2 -4] = -13$$ 5. Simplify constants: $$(x-1)^2 - 1 + 4(y-2)^2 - 16 = -13$$ $$(x-1)^2 + 4(y-2)^2 - 17 = -13$$ 6. Add 17 to both sides: $$(x-1)^2 + 4(y-2)^2 = 4$$ 7. Divide by 4: $$\frac{(x-1)^2}{4} + (y-2)^2 = 1$$ **a. Standard equation:** $$\frac{(x-1)^2}{4} + \frac{(y-2)^2}{1} = 1$$ **b. Vertices:** Center at $(1,2)$ $a=2$ along $x$ (major axis), vertices: $(1 \pm 2, 2) = (-1, 2)$ and $(3, 2)$ **c. Co-vertices:** $b=1$ along $y$, co-vertices: $(1, 2 \pm 1) = (1, 1)$ and $(1, 3)$ **d. Foci:** $$c^2 = a^2 - b^2 = 4 - 1 = 3$$ $$c=\sqrt{3}$$ Foci at $(1 \pm \sqrt{3}, 2)$ **e. Length of axes:** Major axis: $2a=4$ Minor axis: $2b=2$ **f. Asymptotes:** Not applicable to ellipse **g. Eccentricity:** $$e = \frac{c}{a} = \frac{\sqrt{3}}{2}$$ --- **B. Given: 690°** 1. **Draw the angle and label its initial and terminal sides:** - Initial side lies along 0° (positive x-axis). - Terminal side corresponds to angle $690^\circ$. - Since $690^\circ = 690 - 360\times1 = 330^\circ$ (one full rotation subtracted), terminal side same as $330^\circ$. 2. **Find the reference angle:** Reference angle = difference between terminal side and nearest x-axis. $$360^\circ - 330^\circ = 30^\circ$$ 3. **Find the coordinates of the angle:** Coordinates on unit circle for $330^\circ$: $$\left(\cos 330^\circ, \sin 330^\circ\right) = \left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$ 4. **Find exact values of the six trig functions:** $$\sin 690^\circ = \sin 330^\circ = -\frac{1}{2}$$ $$\cos 690^\circ = \cos 330^\circ = \frac{\sqrt{3}}{2}$$ $$\tan 690^\circ = \tan 330^\circ = \frac{\sin}{\cos} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$ $$\csc 690^\circ = \frac{1}{\sin 690^\circ} = -2$$ $$\sec 690^\circ = \frac{1}{\cos 690^\circ} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$ $$\cot 690^\circ = \frac{1}{\tan 690^\circ} = -\sqrt{3}$$