Chapter 6: Problem 37
Find the vertex, focus, and directrix of the parabola, and sketch its graph. \(x^{2}+6 y=0\)
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Chapter 6: Problem 37
Find the vertex, focus, and directrix of the parabola, and sketch its graph. \(x^{2}+6 y=0\)
These are the key concepts you need to understand to accurately answer the question.
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Find an equation of the tangent line to the parabola at the given point, and find the \(x\) -intercept of the line. \(y=-2 x^{2},(2,-8)\)
Determine whether the statement is true or false. Justify your answer. If the vertex and focus of a parabola are on a horizontal line, then the directrix of the parabola is vertical.
Identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. \(\frac{x^{2}}{4 / 9}+\frac{(y+1)^{2}}{4 / 9}=1\)
Find the standard form of the equation of the ellipse with the given characteristics and center at the origin. Vertices: (0,±5) ; passes through the point (4,2)
A simply supported beam is 12 meters long and has a load at the center (see figure). The deflection of the beam at its center is 2 centimeters. Assume that the shape of the deflected beam is parabolic. (a) Write an equation of the parabola. (Assume that the origin is at the center of the deflected beam.) (b) How far from the center of the beam is the deflection equal to 1 centimeter?
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