1. The problem asks to analyze and solve parts related to the conic sections given by the equations: (1) $9x^2 - 16y^2 - 36x - 32y - 124 = 0$ (a homogeneous line equation) (2) $\frac{x^2}{5} + \frac{y^2}{1} = 1$ (an ellipse) (3) Find the coordinates of the ellipse's foci given the axes. (4) From the parabola equation $3y^2 - 10x - 12y - 18 = 0$, find the length of the ellipse's latus rectum. --- 2. **Step 1: Rewrite the first equation in standard form** Given: $$9x^2 - 16y^2 - 36x - 32y - 124 = 0$$ Group $x$ and $y$ terms: $$9x^2 - 36x - 16y^2 - 32y = 124$$ Complete the square for $x$ and $y$: For $x$: $$9(x^2 - 4x) = 9[(x^2 - 4x + 4) - 4] = 9(x - 2)^2 - 36$$ For $y$: $$-16(y^2 + 2y) = -16[(y^2 + 2y + 1) - 1] = -16(y + 1)^2 + 16$$ Rewrite the equation: $$9(x - 2)^2 - 36 - 16(y + 1)^2 + 16 = 124$$ Simplify constants: $$9(x - 2)^2 - 16(y + 1)^2 - 20 = 124$$ Add 20 to both sides: $$9(x - 2)^2 - 16(y + 1)^2 = 144$$ Divide both sides by 144: $$\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1$$ This is the standard form of a hyperbola centered at $(2, -1)$. --- 3. **Step 2: Identify the ellipse center from the second equation** Given ellipse: $$\frac{x^2}{5} + \frac{y^2}{1} = 1$$ This ellipse is centered at the origin $(0,0)$ because there are no linear $x$ or $y$ terms. --- 4. **Step 3: Find the foci of the ellipse** For ellipse equation: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $a^2 = 5$ and $b^2 = 1$. Since $a^2 > b^2$, the major axis is along the $x$-axis. The focal distance $c$ is given by: $$c = \sqrt{a^2 - b^2} = \sqrt{5 - 1} = \sqrt{4} = 2$$ So the foci are at: $$(\pm c, 0) = (\pm 2, 0)$$ --- 5. **Step 4: Find the coordinates of the ellipse's axes intercepts** - $x$-intercepts: Set $y=0$ in ellipse equation: $$\frac{x^2}{5} = 1 \Rightarrow x^2 = 5 \Rightarrow x = \pm \sqrt{5}$$ - $y$-intercepts: Set $x=0$: $$\frac{y^2}{1} = 1 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1$$ So the ellipse intercepts are at $(\pm \sqrt{5}, 0)$ and $(0, \pm 1)$. --- 6. **Step 5: Find the length of the latus rectum of the parabola** Given parabola: $$3y^2 - 10x - 12y - 18 = 0$$ Rewrite in standard form: Group $y$ terms: $$3y^2 - 12y = 10x + 18$$ Complete the square for $y$: $$3(y^2 - 4y) = 10x + 18$$ $$3[(y^2 - 4y + 4) - 4] = 10x + 18$$ $$3(y - 2)^2 - 12 = 10x + 18$$ Add 12 to both sides: $$3(y - 2)^2 = 10x + 30$$ Divide both sides by 10: $$x = \frac{3}{10}(y - 2)^2 - 3$$ This is a parabola opening rightwards with vertex at $(-3, 2)$. The standard form for a parabola opening right is: $$x = a(y - k)^2 + h$$ where $a = \frac{3}{10}$. The length of the latus rectum $L$ is: $$L = \frac{1}{|a|} = \frac{1}{3/10} = \frac{10}{3}$$ --- **Final answers:** - (1) The conic is a hyperbola centered at $(2, -1)$ with equation: $$\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1$$ - (2) The ellipse center is at $(0,0)$. - (3) The ellipse foci are at $(\pm 2, 0)$. - (4) The length of the latus rectum of the parabola is $\frac{10}{3}$.