Exam 10: Analytic Geometry

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Write the word or phrase that best completes each statement or answers the question. - 9x2+49y2=11,0259 x ^ { 2 } + 49 y ^ { 2 } = 11,025

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Determine the two equations necessary to graph the horizontal parabola using a graphing calculator. - x−2=(y−8)2x - 2 = ( y - 8 ) ^ { 2 }

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Identify the type of graph. - −x2−2x+4y2+8y=20- x ^ { 2 } - 2 x + 4 y ^ { 2 } + 8 y = 20

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Give the focus, directrix, and axis for the parabola. - x=9y2x = 9 y ^ { 2 }

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Give the focus, directrix, and axis for the parabola. - (y+3)2=4(x+2)( y + 3 ) ^ { 2 } = 4 ( x + 2 )

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Find the eccentricity of the hyperbola. - x2−25y2=25x ^ { 2 } - 25 y ^ { 2 } = 25

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Give the focus, directrix, and axis for the parabola. - (y−5)2=−12(x+2)( y - 5 ) ^ { 2 } = - 12 ( x + 2 )

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Solve the problem. -An elliptical riding path is to be built on a rectangular piece of property that measures 10 mi by 4 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 Sides Solve the problem. -An elliptical riding path is to be built on a rectangular piece of property that measures 10 mi by 4 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 Sides

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Write the word or phrase that best completes each statement or answers the question. - x216=1−y236\frac { x ^ { 2 } } { 16 } = 1 - \frac { y ^ { 2 } } { 36 }

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Graph the hyperbola. - x225−y29=1\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 9 } = 1  Graph the hyperbola. - \frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 9 } = 1

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Graph the ellipse. - x236+y249=1\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 49 } = 1  Graph the ellipse. - \frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 49 } = 1

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Identify the type of graph. - (x−7)24+(y−6)29=0\frac { ( x - 7 ) ^ { 2 } } { 4 } + \frac { ( y - 6 ) ^ { 2 } } { 9 } = 0

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Identify the equation as a parabola, circle, ellipse, or hyperbola. - y2=144−x2y ^ { 2 } = 144 - x ^ { 2 }

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Write the word or phrase that best completes each statement or answers the question. -To graph x216−y236=1\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 36 } = 1 on a graphics calculator, we must consider the union of the graphs of the two functions, y1=6x216−1y _ { 1 } = 6 \sqrt { \frac { x ^ { 2 } } { 16 } - 1 } and y2=−6x216−1y _ { 2 } = - 6 \sqrt { \frac { x ^ { 2 } } { 16 } - 1 } . Using the graph of y=x216−1y = \frac { x ^ { 2 } } { 16 } - 1 , explain (a) how the solution set of x216−1≥0\frac { x ^ { 2 } } { 16 } - 1 \geq 0 can be determined graphically and (b) how it relates to the domain of the hyperbola.  Write the word or phrase that best completes each statement or answers the question. -To graph  \frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 36 } = 1  on a graphics calculator, we must consider the union of the graphs of the two functions,  y _ { 1 } = 6 \sqrt { \frac { x ^ { 2 } } { 16 } - 1 }  and  y _ { 2 } = - 6 \sqrt { \frac { x ^ { 2 } } { 16 } - 1 } . Using the graph of  y = \frac { x ^ { 2 } } { 16 } - 1 , explain (a) how the solution set of  \frac { x ^ { 2 } } { 16 } - 1 \geq 0  can be determined graphically and (b) how it relates to the domain of the hyperbola.

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Graph the parabola. -Graph the parabola. -

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Identify the equation as a parabola, circle, ellipse, or hyperbola. - 4x=3y2−314 x = 3 y ^ { 2 } - 31

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Match the equation of the parabola with the appropriate description. - x=2y2−6y+2x = 2 y ^ { 2 } - 6 y + 2

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Solve the problem. -A railroad tunnel is shaped like a semi-ellipse. The height of the tunnel at the center is 18 ft and the vertical clearance must be 12 ft at a point 5 ft from the center. Find an equation for the ellipse. Solve the problem. -A railroad tunnel is shaped like a semi-ellipse. The height of the tunnel at the center is 18 ft and the vertical clearance must be 12 ft at a point 5 ft from the center. Find an equation for the ellipse.

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Identify the equation as a parabola, circle, ellipse, or hyperbola. - 16x2+9y2=14416 x ^ { 2 } + 9 y ^ { 2 } = 144

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Determine the two equations necessary to graph the ellipse with a graphing calculator. - (x−3)281+y264=1\frac { ( x - 3 ) ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 64 } = 1

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