### Problem A1 Given the equation: $$y^2 - 25x^2 + 4y + 50x - 46 = 0$$ 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 back: $$(y + 2)^2 - 4 - 25(x - 1)^2 + 25 = 46$$ 5. Simplify: $$(y + 2)^2 - 25(x - 1)^2 + 21 = 46$$ $$(y + 2)^2 - 25(x - 1)^2 = 25$$ 6. Divide both sides by 25: $$\frac{(y + 2)^2}{25} - (x - 1)^2 = 1$$ **Standard equation:** $$\frac{(y + 2)^2}{25} - (x - 1)^2 = 1$$ 7. This is a hyperbola centered at $(1, -2)$ with vertical transverse axis. 8. Vertices: Move $a=5$ units along $y$ from center: $$ (1, -2 \pm 5) = (1, 3) \text{ and } (1, -7)$$ 9. Co-vertices: Move $b=1$ unit along $x$ from center: $$ (1 \pm 1, -2) = (0, -2) \text{ and } (2, -2)$$ 10. Foci: Use $c^2 = a^2 + b^2 = 25 + 1 = 26$, so $c = \sqrt{26}$ $$ (1, -2 \pm \sqrt{26})$$ 11. Length of major (transverse) axis: $2a = 10$ 12. Length of minor (conjugate) axis: $2b = 2$ 13. Equations of asymptotes for vertical transverse axis: $$ y = -2 \pm \frac{a}{b}(x -1) = -2 \pm 5(x - 1)$$ 14. Eccentricity: $$e = \frac{c}{a} = \frac{\sqrt{26}}{5}$$ --- ### Problem A2 Given: $$(x^2 / 144) - (y^2 / 49) = 1$$ This is a hyperbola centered at origin, horizontal transverse axis. - $a^2 = 144 \Rightarrow a=12$ - $b^2=49 \Rightarrow b=7$ - Vertices: $( \pm 12, 0)$ - Co-vertices: $(0, \pm 7)$ - Foci: $c^2 = a^2 + b^2 = 144 + 49 = 193$, $c=\sqrt{193}$, foci at $(\pm \sqrt{193},0)$ - Major axis length: $2a=24$ - Minor axis length: $2b=14$ - Asymptotes: $y = \pm \frac{b}{a}x = \pm \frac{7}{12}x$ - Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{193}}{12}$ --- ### Problem A3 Given: $$(x^2/9) + (y^2/25) =1$$ Ellipse centered at origin, vertical major axis because $25>9$: - $a^2=25, a=5$ - $b^2=9, b=3$ - Vertices: $(0, \pm 5)$ - Co-vertices: $(\pm 3, 0)$ - Foci: $c^2 = a^2 - b^2 = 25 - 9 = 16$, $c=4$ - Foci at $(0, \pm 4)$ - Major axis length: $2a=10$ - Minor axis length: $2b=6$ - Eccentricity: $e = \frac{c}{a} = \frac{4}{5}$ - Ellipses do not have asymptotes --- ### Problem A4 Given: $$x^2 - 2x + 4y^2 - 16y = -13$$ 1. Complete the square for $x$: $$x^2 - 2x = (x-1)^2 -1$$ 2. Complete square for $y$: $$4y^2 -16y = 4(y^2 -4y) = 4[(y-2)^2 -4] = 4(y-2)^2 -16$$ 3. Substitute: $$(x-1)^2 -1 + 4(y-2)^2 -16 = -13$$ 4. Simplify: $$(x-1)^2 + 4(y-2)^2 -17 = -13$$ $$(x-1)^2 + 4(y-2)^2 = 4$$ 5. Divide by 4: $$\frac{(x-1)^2}{4} + (y-2)^2 =1$$ This is an ellipse centered at $(1, 2)$ with horizontal major axis. - Compare: $a^2=4$, $b^2=1$ - $a=2$, $b=1$ - Vertices: $(1 \pm 2, 2) = (-1, 2)$ and $(3, 2)$ - Co-vertices: $(1, 2 \pm 1) = (1, 1)$ and $(1, 3)$ - Foci: $c^2 = a^2 - b^2 = 4 -1=3$, $c=\sqrt{3}$ Foci at $(1 \pm \sqrt{3}, 2)$ - Major axis length: $2a=4$ - Minor axis length: $2b=2$ - No asymptotes - Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{3}}{2}$ --- ### Problem B Given angle: $690^\circ$ 1. Find equivalent angle modulo 360: $$690^\circ - 360^\circ = 330^\circ$$ 2. Reference angle: $$360^\circ - 330^\circ = 30^\circ$$ 3. Coordinates at $330^\circ$ (unit circle): $$(\cos 330^\circ, \sin 330^\circ) = \left(\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$$ 4. Six trigonometric function values: - $$\sin 690^\circ = \sin 330^\circ = -\frac{1}{2}$$ - $$\cos 690^\circ = \cos 330^\circ = \frac{\sqrt{3}}{2}$$ - $$\tan 690^\circ = \tan 330^\circ = \frac{\sin}{\cos} = -\frac{1}{2} \div \frac{\sqrt{3}}{2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$$ - $$\cot 690^\circ = \frac{1}{\tan 690^\circ} = -\sqrt{3}$$ - $$\sec 690^\circ = \frac{1}{\cos 690^\circ} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$ - $$\csc 690^\circ = \frac{1}{\sin 690^\circ} = -2$$ --- Final answers are presented with clear intermediate steps for all parts.