Exam 9: Analytic Geometry

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Choose the one alternative that best completes the statement or answers the question. Use a graphing utility to graph the curve defined by the given parametric equations. - x=2t2,y=t+2;−∞<t<∞x=2 t^{2}, y=t+2 ;-\infty<t<\infty  Choose the one alternative that best completes the statement or answers the question. Use a graphing utility to graph the curve defined by the given parametric equations. - x=2 t^{2}, y=t+2 ;-\infty<t<\infty

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Find an equation of the parabola described. -Focus at (0, 21); directrix the line y = -21

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Write the word or phrase that best completes each statement or answers the question. The parametric equations of four curves are given. Graph each of them, indicating the orientation. - :x=7t,y=7-7t;\leqt\leq :x=t,y=;\leqt\leq :x=-8,y=t-3;-4\leqt\leq4 :x=t-5,y=t+2;-4\leqt\leq7

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Find an equation for the hyperbola described. -Vertices (12,−3)\left( \frac { 1 } { 2 } , - 3 \right) and (−92,−3);\left( - \frac { 9 } { 2 } , - 3 \right) ; asymptotes y+3=±65(x+2)y + 3 = \pm \frac { 6 } { 5 } ( x + 2 )

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Find an equation of the parabola described. -Focus at (4, 0); vertex at (0, 0)

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Find an equation for the ellipse described. Graph the equation. -Center at (−2,3);( - 2,3 ) ; focus at (−7,3);( - 7,3 ) ; contains the point (−8,3)( - 8,3 )  Find an equation for the ellipse described. Graph the equation. -Center at  ( - 2,3 ) ;  focus at  ( - 7,3 ) ;  contains the point  ( - 8,3 )

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Find an equation for the parabola described. -Vertex at (6,1)( 6,1 ) ; focus at (6,3)( 6,3 )

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Solve the problem. -An experimental model for a suspension bridge is built in the shape of a parabolic arch. In one section, cable runs from the top of one tower down to the roadway, just touching it there, and up again to the top of a second Tower. The towers are both 12.25 inches tall and stand 70 inches apart. At some point along the road from the Lowest point of the cable, the cable is 1.96 inches above the roadway. Find the distance between that point and The base of the nearest tower.

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Write the word or phrase that best completes each statement or answers the question. -A sealed-beam headlight is in the shape of a paraboloid of revolution. The bulb, which is placed at the focus, is 3 centimeters from the vertex. If the depth is to be 6 centimeters, what is the diameter of the headlight at its opening?

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Graph the hyperbola. - 25y2=9x2+22525 y^{2}=9 x^{2}+225  Graph the hyperbola. - 25 y^{2}=9 x^{2}+225

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Identify the equation without applying a rotation of axes. - 2x2+8xy+16y2−3x−2y+10=02 x ^ { 2 } + 8 x y + 16 y ^ { 2 } - 3 x - 2 y + 10 = 0

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Identify the equation without completing the square. - 3x2+4y2+4x+2y=03 x ^ { 2 } + 4 y ^ { 2 } + 4 x + 2 y = 0

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Find an equation for the ellipse described. -Center at (6,4)( 6,4 ) ; focus at (10,4)( 10,4 ) ; vertex at (12,4)( 12,4 )

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Identify the conic that the polar equation represents. Also, give the position of the directrix. - r=93+3sin⁡θr = \frac { 9 } { 3 + 3 \sin \theta }

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Graph the curve whose parametric equations are given. - x=3cos⁡t,y=−3sin⁡t;π2≤t≤3π2x = 3 \cos t , y = - 3 \sin t ; \frac { \pi } { 2 } \leq t \leq \frac { 3 \pi } { 2 }  Graph the curve whose parametric equations are given. - x = 3 \cos t , y = - 3 \sin t ; \frac { \pi } { 2 } \leq t \leq \frac { 3 \pi } { 2 }

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Solve. -A baseball pitcher throws a baseball with an initial speed of 140 feet per second at an angle of 20∘20 ^ { \circ } to the horizontal. The ball leaves the pitcher's hand at a height of 5 feet. Find parametric equations that describe the motion of the ball as a function of time. How long is the ball in the air? When is the ball at its maximum height? What is the maximum height of the ball?

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Convert the polar equation to a rectangular equation. - r=4sec⁡θ4sec⁡θ+1r = \frac { 4 \sec \theta } { 4 \sec \theta + 1 }

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Write an equation for the parabola. -Write an equation for the parabola. -

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Find an equation for the ellipse described. -Focus at (−3,0)( - 3,0 ) ; vertices at (±5,0)( \pm 5,0 )

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Rotate the axes so that the new equation contains no xy-term. Discuss the new equation. - 31x2+103xy+21y2−144=031 x^{2}+10 \sqrt{3} x y+21 y^{2}-144=0

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