1. Write the equation of the hyperbola with center $(4,-3)$, horizontal transverse axis length $5$, conjugate axis length $4$. Step 1: Center $(h,k)=(4,-3)$, transverse axis length $2a=5\implies a=\frac{5}{2}$. Conjugate axis length $2b=4\implies b=2$. Step 2: Horizontal transverse axis means equation: $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} =1$$ $$\Rightarrow \frac{(x-4)^2}{(\frac{5}{2})^2} - \frac{(y+3)^2}{2^2} =1$$ $$\Rightarrow \frac{(x-4)^2}{\frac{25}{4}} - \frac{(y+3)^2}{4} =1$$ Step 3: Find $c$ for foci: $$c^2 = a^2 + b^2 = \left( \frac{25}{4} \right) + 4 = \frac{25}{4} + \frac{16}{4} = \frac{41}{4}$$ $$c = \frac{\sqrt{41}}{2}$$ Step 4: Foci coordinates (horizontal transverse axis): $$(h\pm c,k) = \left(4 \pm \frac{\sqrt{41}}{2},-3\right)$$ Step 5: Vertices at $$(h\pm a, k) = \left(4 \pm \frac{5}{2}, -3 \right)$$ Step 6: Endpoints of transverse axis are vertices: $\left(4 \pm \frac{5}{2},-3\right)$. Endpoints of conjugate axis: $$(h, k \pm b) = \left(4, -3 \pm 2 \right)$$ Step 7: Length of latus rectum = $\frac{2b^2}{a} = \frac{2 \cdot 4}{\frac{5}{2}}= \frac{8}{\frac{5}{2}} = \frac{8 \cdot 2}{5} = \frac{16}{5}$ Coordinates of latus rectum endpoints are at distance $\frac{16}{5}/2=\frac{8}{5}$ above and below each focus: For each focus $(x_f,y_f)$: endpoints $$(x_f, y_f \pm \frac{8}{5})$$ --- 2. Foci at $(-3,-1)$ and $(7,-1)$, conjugate axis length $6$. Step 1: Center is midpoint: $$\left( \frac{-3+7}{2}, \frac{-1-1}{2} \right) = (2,-1)$$ Step 2: Distance between foci = $2c = \sqrt{(7+3)^2 + (-1+1)^2} = \sqrt{10^2+0} = 10$ so $$c = 5$$ Step 3: Since foci have same $y$, transverse axis is horizontal. Step 4: Conjugate axis length $2b=6 \implies b=3$. Step 5: Use $c^2 = a^2 + b^2$ to find $a$: $$25 = a^2 + 9 \Rightarrow a^2 = 16 \Rightarrow a = 4$$ Step 6: Equation: $$\frac{(x-2)^2}{16} - \frac{(y+1)^2}{9} = 1$$ Step 7: Vertices: $$(h \pm a, k) = (2 \pm 4, -1) = (-2, -1), (6,-1)$$ Step 8: Endpoints of transverse axis = vertices Endpoints of conjugate axis: $$(h, k \pm b) = (2, -1 \pm 3) = (2, 2), (2,-4)$$ Step 9: Length latus rectum: $$\frac{2b^2}{a} = \frac{2 \cdot 9}{4} = \frac{18}{4} = \frac{9}{2}$$ Coordinates of latus rectum endpoints are $$(x_f, y_f \pm \frac{9}{4})$$ for each focus $(x_f,y_f)$. --- 3. Find foci, vertices, transverse and conjugate axis endpoints, and latus rectum for $$9x^2 - 4y^2 - 36x + 8y - 4 = 0$$ Step 1: Rewrite grouping $x$ and $y$ terms: $$9x^2 -36x - 4y^2 + 8y =4$$ Step 2: Complete square for $x$ and $y$: $$9(x^2 -4x) -4(y^2 - 2y) =4$$ For $x$: $$x^2 - 4x + 4 -4 = (x-2)^2 -4$$ For $y$: $$y^2 - 2y + 1 -1 = (y-1)^2 -1$$ Step 3: Substitute: $$9[(x-2)^2 -4] -4[(y-1)^2 -1] = 4$$ Step 4: Expand: $$9(x-2)^2 -36 -4(y-1)^2 +4 = 4$$ Step 5: Simplify: $$9(x-2)^2 -4(y-1)^2 -32 = 4$$ Step 6: Add 32 to both sides: $$9(x-2)^2 -4(y-1)^2 = 36$$ Step 7: Divide both sides by 36: $$\frac{(x-2)^2}{4} - \frac{(y-1)^2}{9} = 1$$ Step 8: Here $a^2=4$, $b^2=9$. Since positive term is $x$, transverse axis is horizontal. Step 9: Find $c$: $$c^2 = a^2 + b^2 = 4 + 9 = 13 \Rightarrow c = \sqrt{13}$$ Step 10: Center $(h,k) = (2,1)$. Vertices: $$(h \pm a, k) = (2 \pm 2, 1) = (0,1), (4,1)$$ Foci: $$(2 \pm \sqrt{13}, 1)$$ Endpoints conjugate axis: $$(2, 1 \pm 3) = (2,4), (2,-2)$$ Length latus rectum: $$\frac{2b^2}{a} = \frac{2 \cdot 9}{2} = 9$$ Coordinates of latus rectum endpoints: For each focus $(x_f,y_f)$, $$(x_f, y_f \pm \frac{9}{4})$$ --- 4. Find foci, vertices, axes endpoints, latus rectum for $$x^2 - 4y^2 + 2x -16y -23 = 0$$ Step 1: Group $x$ and $y$: $$x^2 + 2x -4(y^2 + 4y) = 23$$ Step 2: Complete square: $$x^2 + 2x + 1 - 1 - 4(y^2 +4y + 4 -4) = 23$$ Rewrite: $$(x+1)^2 -1 -4[(y+2)^2 -4] = 23$$ Step 3: Expand: $$(x+1)^2 -1 -4(y+2)^2 + 16 = 23$$ Step 4: Simplify constants: $$(x+1)^2 -4(y+2)^2 + 15 = 23$$ $$ (x+1)^2 -4(y+2)^2 = 8$$ Step 5: Divide both sides by 8: $$\frac{(x+1)^2}{8} - \frac{(y+2)^2}{2} = 1$$ Step 6: Here $a^2 =8$, $b^2 = 2$, transverse axis horizontal. Step 7: Calculate $c$: $$c^2 = a^2 + b^2 = 8 + 2 = 10 \Rightarrow c = \sqrt{10}$$ Step 8: Center $(h,k) = (-1, -2)$. Vertices: $$(h \pm a, k) = \left(-1 \pm \sqrt{8}, -2\right) = \left(-1 \pm 2\sqrt{2}, -2\right)$$ Foci: $$( -1 \pm \sqrt{10}, -2 )$$ Endpoints of conjugate axis: $$( -1, -2 \pm \sqrt{2} )$$ Length latus rectum: $$\frac{2b^2}{a} = \frac{2 \cdot 2}{\sqrt{8}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$ Latus rectum endpoints: $$(x_f, y_f \pm \frac{\sqrt{2}}{2})$$ --- 5. Equation $$25x^2 -4y^2 + 50x + 8y + 121 = 0$$ Step 1: Group terms: $$25x^2 + 50x -4y^2 + 8y = -121$$ Step 2: Complete square: $$25(x^2 + 2x) - 4(y^2 - 2y) = -121$$ Step 3: Complete inside parentheses: $$x^2 + 2x + 1 -1, y^2 - 2y + 1 - 1$$ $$25[(x+1)^2 -1] -4[(y-1)^2 - 1] = -121$$ Step 4: Expand: $$25(x+1)^2 - 25 -4(y-1)^2 + 4 = -121$$ Step 5: Simplify constants: $$25(x+1)^2 - 4(y-1)^2 - 21 = -121$$ $$25(x+1)^2 - 4(y-1)^2 = -100$$ Step 6: Divide both sides by -100: $$\frac{(y-1)^2}{25} - \frac{(x+1)^2}{4} = 1$$ Step 7: Now transverse axis is vertical, since $y$ term positive. Step 8: Identify $a^2 = 25$, $b^2 = 4$, center $(h,k) = (-1,1)$. Step 9: Calculate $c$: $$c^2 = a^2 + b^2 = 25 + 4 = 29$$ $$c = \sqrt{29}$$ Step 10: Vertices: $$(h, k \pm a) = (-1, 1 \pm 5) = (-1, 6), (-1, -4)$$ Foci: $$(h, k \pm c) = (-1, 1 \pm \sqrt{29})$$ Endpoints conjugate axis: $$(h \pm b, k) = (-1 \pm 2, 1) = (1,1), (-3,1)$$ Length latus rectum: $$\frac{2b^2}{a} = \frac{2 \cdot 4}{5} = \frac{8}{5}$$ Latus rectum endpoints for each focus $(h,y_f)$: $$(h \pm \frac{8}{10}, y_f) = \left(-1 \pm \frac{4}{5}, 1 \pm \sqrt{29} \right)$$ --- "Final Answers Summary": 1. Equation: $$\frac{(x-4)^2}{\frac{25}{4}} - \frac{(y+3)^2}{4} = 1$$ Foci: $$\left(4 \pm \frac{\sqrt{41}}{2}, -3\right)$$ Vertices: $$\left(4 \pm \frac{5}{2}, -3\right)$$ Conjugate axis endpoints: $$(4, -3 \pm 2)$$ Latus rectum length: $$\frac{16}{5}$$ 2. Equation: $$\frac{(x-2)^2}{16} - \frac{(y+1)^2}{9} = 1$$ Foci: $$(2 \pm 5, -1)$$ Vertices: $$(2 \pm 4, -1)$$ Conjugate axis endpoints: $$(2, -1 \pm 3)$$ Latus rectum length: $$\frac{9}{2}$$ 3. Equation: $$\frac{(x-2)^2}{4} - \frac{(y-1)^2}{9} = 1$$ Foci: $$(2 \pm \sqrt{13}, 1)$$ Vertices: $$(0,1), (4,1)$$ Conjugate axis endpoints: $$(2,4), (2,-2)$$ Latus rectum length: $$9$$ 4. Equation: $$\frac{(x+1)^2}{8} - \frac{(y+2)^2}{2} = 1$$ Foci: $$(-1 \pm \sqrt{10}, -2)$$ Vertices: $$(-1 \pm 2\sqrt{2}, -2)$$ Conjugate axis endpoints: $$(-1, -2 \pm \sqrt{2})$$ Latus rectum length: $$\sqrt{2}$$ 5. Equation: $$\frac{(y-1)^2}{25} - \frac{(x+1)^2}{4} = 1$$ Foci: $$(-1, 1 \pm \sqrt{29})$$ Vertices: $$(-1, 1 \pm 5)$$ Conjugate axis endpoints: $$(-1 \pm 2, 1)$$ Latus rectum length: $$\frac{8}{5}$$