Question: Midterm Quiz #2 (10 pts each)\n\nFind the center, vertices, foci for each of the following conic section and sketch the graph.\n\n1. 16x^2 + 25y^2 + 16x + 4 = 100y\n\n2. 9x^2 - 16y^2 + 18x + 64y = 91\n\nIn each of the following, find the equation of the circle determined by the given conditions.\n\n3. Has radius 5 and tangent to the line 3x + 4y = 24 at the point (2, 4½).\n\n4. Circumscribes the triangle determined by the lines y = 0, y = x, and 2x + 3y = 10.\n\nIn each of the following, find the equation of the ellipse determined by the given conditions.\n\n5. Center (1, 3), vertex (1, −1), and passing through the origin.\n\n6. Center (2, 0), focus (2, 6), eccentricity 2/3.\n\nIn each of the following, find the equation of the parabola determined by the given conditions.\n\n7. Axis vertical, latus rectum 4, and passing through (6, 4) and (−4, −1).\n\n8. Focus (−1,4), end point of latus rectum (−1,1).\n\nIn each of the following, find the equation of the hyperbola determined by the given conditions.\n\n9. Asymptotes x + y = 1 and x − y = 1, and passing through (3, 1).\n\n10. Center (2, 0), focus (2, 4), eccentricity √2.
1. **Problem:** Find the center, vertices, and foci of the conic given by $$16x^2 + 25y^2 + 16x + 4 = 100y$$.\n\n2. **Rewrite the equation:** Move all terms to one side:\n$$16x^2 + 25y^2 + 16x + 4 - 100y = 0$$\n\n3. **Group x and y terms:**\n$$16x^2 + 16x + 25y^2 - 100y + 4 = 0$$\n\n4. **Complete the square for x and y:**\n- For x: $$16(x^2 + x)$$\n- For y: $$25(y^2 - 4y)$$\n\n5. **Complete the square inside parentheses:**\n- $$x^2 + x = x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 = (x + \frac{1}{2})^2 - \frac{1}{4}$$\n- $$y^2 - 4y = y^2 - 4y + 4 - 4 = (y - 2)^2 - 4$$\n\n6. **Substitute back:**\n$$16\left((x + \frac{1}{2})^2 - \frac{1}{4}\right) + 25\left((y - 2)^2 - 4\right) + 4 = 0$$\n\n7. **Expand:**\n$$16(x + \frac{1}{2})^2 - 16 \times \frac{1}{4} + 25(y - 2)^2 - 100 + 4 = 0$$\n$$16(x + \frac{1}{2})^2 + 25(y - 2)^2 - 4 - 100 + 4 = 0$$\n$$16(x + \frac{1}{2})^2 + 25(y - 2)^2 - 100 = 0$$\n\n8. **Add 100 to both sides:**\n$$16(x + \frac{1}{2})^2 + 25(y - 2)^2 = 100$$\n\n9. **Divide both sides by 100:**\n$$\frac{16}{100}(x + \frac{1}{2})^2 + \frac{25}{100}(y - 2)^2 = 1$$\n\n10. **Simplify fractions:**\n$$\frac{(x + \frac{1}{2})^2}{\frac{100}{16}} + \frac{(y - 2)^2}{\frac{100}{25}} = 1$$\n$$\frac{(x + \frac{1}{2})^2}{6.25} + \frac{(y - 2)^2}{4} = 1$$\n\n11. **Identify ellipse parameters:**\n- Center: $$\left(-\frac{1}{2}, 2\right)$$\n- $$a^2 = 6.25$$, $$a = 2.5$$ (larger denominator)\n- $$b^2 = 4$$, $$b = 2$$\n\n12. **Vertices:** Since $$a^2 > b^2$$ and $$a^2$$ is under x-term, major axis is horizontal.\n- Vertices: $$\left(-\frac{1}{2} \pm 2.5, 2\right) = (2, 2) \text{ and } (-3, 2)$$\n\n13. **Foci:** Calculate $$c$$ where $$c^2 = a^2 - b^2 = 6.25 - 4 = 2.25$$, so $$c = 1.5$$\n- Foci: $$\left(-\frac{1}{2} \pm 1.5, 2\right) = (1, 2) \text{ and } (-2, 2)$$\n\n**Final answer:**\n- Center: $$\left(-\frac{1}{2}, 2\right)$$\n- Vertices: $$(2, 2) \text{ and } (-3, 2)$$\n- Foci: $$(1, 2) \text{ and } (-2, 2)$$