Exam 60: Introduction to Conics Parabolas

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Find the vertex and focus of the parabola. y2=92xy ^ { 2 } = - \frac { 9 } { 2 } x

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Find the standard form of the equation of the ellipse with the following graph. Find the standard form of the equation of the ellipse with the following graph.

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Give the coordinates of the circle's center and its radius.​ (x6)2+(y+5)2=4( x - 6 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 4

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A simply supported beam is 12 meters long and has a load at the center (see figure).​  A simply supported beam is 12 meters long and has a load at the center (see figure).​   ​ where,  a = 12 , b = 2  The deflection of the beam at its center is 2 centimeters.Assume that the shape of the deflected beam is parabolic.Write an equation of the parabola.(Assume that the origin is at the center of the deflected beam. ) ​ ​ where, a=12,b=2a = 12 , b = 2 The deflection of the beam at its center is 2 centimeters.Assume that the shape of the deflected beam is parabolic.Write an equation of the parabola.(Assume that the origin is at the center of the deflected beam. ) ​

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Select the graph of the following equation.​ x2=8yx ^ { 2 } = 8 y

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Find the vertex and focus of the parabola. y2=34xy ^ { 2 } = - \frac { 3 } { 4 } x

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Find the standard form of the equation of the parabola with the given characteristic and vertex at the origin. Focus: (0,12)\left( 0 , \frac { 1 } { 2 } \right)

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Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. ​ Vertices: (0,±2);asymptotes: y = ± 32\frac { 3 } { 2 } x ​

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The revenue R (in dollars)generated by the sale of x units of a patio furniture set is given by (x116)2=45(R16,820)( x - 116 ) ^ { 2 } = - \frac { 4 } { 5 } ( R - 16,820 ) .Approximate the number of sales that will maximize revenue. ​

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Find the standard form of the equation of the parabola with the given characteristic and vertex at the origin. focus: (0,-4)

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Find the vertex,focus,and directrix of the parabola.​ x2+4y=0x ^ { 2 } + 4 y = 0

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Find the standard form of the equation of the hyperbola with the given characteristics. vertices: (0,±4)focies: (0,±5)

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Find the standard form of the equation of the parabola with the given characteristic and vertex at the origin. Focus: (0,4)( 0 , - 4 )

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Find the standard form of the equation of the parabola with the given characteristic and vertex at the origin. directrix: x = -5 ​

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Find the vertex and focus of the parabola. y2=17xy ^ { 2 } = - \frac { 1 } { 7 } x

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Find the center and vertices which located on the major axis of the ellipse from given equation and select its graph.​ x236+y29=1\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 9 } = 1

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Find the vertex,focus,and directrix of the parabola.​ (x1)2+16(y+6)=0( x - 1 ) ^ { 2 } + 16 ( y + 6 ) = 0

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A solar oven uses a parabolic reflector to focus the sun's rays at a point 5 inches from the vertex of the reflector (see figure).Write an equation for a cross section of the oven's reflector with its focus on the positive y axis and its vertex at the origin. A solar oven uses a parabolic reflector to focus the sun's rays at a point 5 inches from the vertex of the reflector (see figure).Write an equation for a cross section of the oven's reflector with its focus on the positive y axis and its vertex at the origin.   L = 5 inches L = 5 inches

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Find the vertex,focus,and directrix of the parabola.​ (x+3)+(y1)2=0( x + 3 ) + ( y - 1 ) ^ { 2 } = 0

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The path of a projectile projected horizontally with a velocity of v feet per second at a height of s feet,where the model for the path is x2=v216(ys)x ^ { 2 } = - \frac { v ^ { 2 } } { 16 } ( y - s ) .In this model (in which air resistance is disregarded),y is the height (in feet)of the projectile and x is the horizontal distance (in feet)the projectile travels.A ball is thrown from the top of a 150-foot tower with a velocity of 32 feet per second.Find the equation of the parabolic path. ​

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