Exam 6: Topics in Analytic Geometry

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Find the center, vertices and foci of the hyperbola. x225y216=1\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 16 } = 1

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Identify the conic and select its correct graph. r=102cosθr = \frac { 10 } { 2 - \cos \theta }

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Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points. r=3r = - 3

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Select the curve represented by the parametric equations. (indicate the orientation of the curve) x=4-5t y=5+4t

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Select the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. x2+y27x+5y17=0x ^ { 2 } + y ^ { 2 } - 7 x + 5 y - 17 = 0

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Find the standard form of the equation of the ellipse with the given characteristics and center at the origin. Find the standard form of the equation of the ellipse with the given characteristics and center at the origin.

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Find the vertex, focus, and directrix of the parabola. (x+5)2=4(y52)( x + 5 ) ^ { 2 } = 4 \left( y - \frac { 5 } { 2 } \right)

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A point in polar coordinates is given. Convert the point to rectangular coordinates. Round your answers to one decimal places. (2.1,1.8)( - 2.1,1.8 )

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Consider a line with slope m and y-intercept (0,2)( 0,2 ) Select the graph of the distance between the origin and the line.

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By using a graphing utility select the correct graph of the polar equation. Identify the graph. 1111+16sinθ\frac { 11 } { 11 + 16 \sin \theta }

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Select the graph of the polar equation using symmetry, zeros, maximum r-values, and any other additional points. r=56cosθr = 5 - 6 \cos \theta

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Use a graphing utility to find the rectangular coordinates of the point given in polar coordinates. Round your results to two decimal places. (6,5π3)\left( 6 , \frac { 5 \pi } { 3 } \right)

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Find the vertex and focus of the parabola. y2=98xy ^ { 2 } = - \frac { 9 } { 8 } x

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Convert the polar equation to rectangular form. r=6cscθr = 6 \csc \theta

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Convert the rectangular equation to polar form. Assume a>0a > 0 xy=17x y = 17

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Convert the rectangular equation to polar form. Assume a>0a > 0 y=6y = - 6

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Select the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. x2+y26x+2y19=0x ^ { 2 } + y ^ { 2 } - 6 x + 2 y - 19 = 0

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Convert the polar equation to rectangular form. r2=4sinθr ^ { 2 } = 4 \sin \theta

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A point in rectangular coordinates is given. Convert the point to polar coordinates. (3,0)( - 3,0 )

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Using following result find a set of parametric equation of the line. x=x1+t(x2x1),y=y1+t(y2y1)x = x _ { 1 } + t \left( x _ { 2 } - x _ { 1 } \right) , y = y _ { 1 } + t \left( y _ { 2 } - y _ { 1 } \right) Line: passes through (0,0)( 0,0 ) and (9,2)( 9,2 )

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