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:Let's analyze and solve the problem step-by-step.---✏️ Given:The curve is symmetric around the y-axis, and centered at (0, 0).Vertices: (−2, 0) and (2, 0)Foci: (−2√17, 0) and (2√17, 0)---✅ Step 1: Identify the type of conic sectionSince:The center is at (0, 0)The vertices and foci lie along the x-axisThe foci are farther than the vertices from the centerThis is a hyperbola that opens left and right (horizontally).---✅ Step 2: Standard form of a horizontal hyperbola\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1Where:a = distance from center to a vertexc = distance from center to a focusb is found from: ---✅ Step 3: Use the given valuesVertices: Foci: Now solve for :c^2 = a^2 + b^2 \\(2\sqrt{17})^2 = 2^2 + b^2 \\4 \cdot 17 = 4 + b^2 \\68 = 4 + b^2 \\b^2 = 64---✅ Step 4: Final equation of the hyperbola\frac{x^2}{4} - \frac{y^2}{64} = 1---✅ Step 5: Directrix of a hyperbolaThe directrix of a horizontally-opening hyperbola is:x = \pm \frac{a^2}{\sqrt{c^2}} = \pm \frac{4}{2\sqrt{17}} = \pm \frac{2}{\sqrt{17}}Or rationalized:x = \pm \frac{2\sqrt{17}}{17}---

Answered by chedreineellardazur | 2025-07-05