Analyze and solve the problem below.
Make an approximate sketch of the curve of the Eiffel Tower on the cartesian plane, with its center at (0,0), and say that the vertices is at (−2, 0), (2, 0), and the foci is at (−2√17, 0), (2√17, 0). Give the type of conic section, and its directrix.
Answer (1)
Step-by-step explanation:Here's an analysis and solution to the problem: 1. Type of Conic Section: Given that the foci are at (−2√17, 0) and (2√17, 0), and the vertices are at (−2, 0) and (2, 0), this conic section is a hyperbola. The vertices and foci lie on the x-axis, indicating a horizontally oriented hyperbola. 2. Equation of the Hyperbola: The general equation for a horizontally oriented hyperbola centered at (0,0) is: $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ Where: - 'a' is the distance from the center to each vertex.- 'c' is the distance from the center to each focus.- b^2 = c^2 - a^2 From the given information: - a = 2 (distance from (0,0) to (2,0) or (-2,0))- c = 2√17 (distance from (0,0) to (2√17,0) or (-2√17,0)) Therefore: b^2 = (2\sqrt{17})^2 - 2^2 = 68 - 4 = 64b = 8 The equation of the hyperbola representing the Eiffel Tower sketch is: $ \frac{x^2}{4} - \frac{y^2}{64} = 1 $ 3. Directrix: The directrix for a horizontally oriented hyperbola is given by: x = \pm \frac{a^2}{c} Substituting the values of 'a' and 'c': x = \pm \frac{2^2}{2\sqrt{17}} = \pm \frac{4}{2\sqrt{17}} = \pm \frac{2}{\sqrt{17}} = \pm \frac{2\sqrt{17}}{17} Therefore, the directrices are: x = \frac{2\sqrt{17}}{17} and x = -\frac{2\sqrt{17}}{17} 4. Approximate Sketch: An approximate sketch would show a hyperbola opening left and right, centered at (0,0). The vertices would be at (−2, 0) and (2, 0). The curve would be wider along the y-axis than along the x-axis, reflecting the larger value of 'b' compared to 'a'. The foci would be further out from the center than the vertices. It would not be a perfect representation of the Eiffel Tower's actual curve, as that curve is far more complex. This is just a simplified mathematical model. The sketch should also indicate the directrices as vertical lines at approximately x = \pm 0.29. In summary: - Type of conic section: Hyperbola- Equation: $ \frac{x^2}{4} - \frac{y^2}{64} = 1 $- Directrices: x = \frac{2\sqrt{17}}{17} and x = -\frac{2\sqrt{17}}{17} (approximately x = ±0.29) Remember that this is a simplified mathematical representation and doesn't perfectly capture the complex curves of the actual Eiffel Tower.
