Exam 12: Parametric Equations and Polar Coordinates

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Graph. - 48x2+27y2=43248 x^{2}+27 y^{2}=432  Graph. - 48 x^{2}+27 y^{2}=432

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Graph the parabola or ellipse. Include the directrix that corresponds to the focus at the origin. - r=2484sinθr = \frac { 24 } { 8 - 4 \sin \theta }  Graph the parabola or ellipse. Include the directrix that corresponds to the focus at the origin. - r = \frac { 24 } { 8 - 4 \sin \theta }

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Find an equation for the line tangent to the curve at the point defined by the given value of t. - x=10t27,y=t3,t=1x = 10 t ^ { 2 } - 7 , y = t ^ { 3 } , t = 1

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Find the area of the specified region. -Inside the smaller loop of the limacon r=4+8sinθr = 4 + 8 \sin \theta

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Solve the problem. -The parabola y2=14xy ^ { 2 } = 14 x is shifted up 7 units and left 7 units. Find an equation for the new parabola and find the new vertex.

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Choose the equation that matches the graph. -Choose the equation that matches the graph. -

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Determine the symmetries of the curve. - r=1+5cosθr = 1 + 5 \cos \theta

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Graph the set of points whose polar coordinates satisfy the given equation or inequality. - r4\mathrm { r } \leq 4  Graph the set of points whose polar coordinates satisfy the given equation or inequality. - \mathrm { r } \leq 4

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Graph the parabola. - x2=2yx^{2}=-2 y  Graph the parabola. - x^{2}=-2 y

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The eccentricity is given of a conic section with one focus at the origin, along with the directrix corresponding to that focus. Find a polar equation for the conic section. -e = 2, x = -8

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 Find a polar equation in the form rcos(θθ0)=r0 for the given line. \text { Find a polar equation in the form } r \cos \left( \theta - \theta _ { 0 } \right) = r _ { 0 } \text { for the given line. } -x = -4

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Provide an appropriate response. -Assuming r=f(θ)\mathrm { r } = \mathrm { f } ( \theta ) is continuous for αθβ\alpha \leq \theta \leq \beta and α<βα+2π\alpha < \beta \leq \alpha + 2 \pi , what can be said about the relative areas betweel the origin and the polar curves and =(\theta),\alpha\leq\theta\leq\beta =2(\theta),\alpha\leq\theta\leq\beta? Give reasons for your answer.

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Solve the problem. -The hyperbola y225x24=1\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 4 } = 1 is shifted horizontally and vertically to obtain the hyperbola (y+5)225(x+3)24=1\frac { ( y + 5 ) ^ { 2 } } { 25 } - \frac { ( x + 3 ) ^ { 2 } } { 4 } = 1 . Graph the new hyperbola.  Solve the problem. -The hyperbola  \frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 4 } = 1  is shifted horizontally and vertically to obtain the hyperbola  \frac { ( y + 5 ) ^ { 2 } } { 25 } - \frac { ( x + 3 ) ^ { 2 } } { 4 } = 1 . Graph the new hyperbola.

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Solve the problem. -The ellipse x236+y24=1\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 4 } = 1 is shifted horizontally and vertically to obtain the ellipse (x1)236+(y+1)24=1.\frac { ( \mathrm { x } - 1 ) ^ { 2 } } { 36 } + \frac { ( \mathrm { y } + 1 ) ^ { 2 } } { 4 } = 1 . Graph the new ellipse.  Solve the problem. -The ellipse  \frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 4 } = 1  is shifted horizontally and vertically to obtain the ellipse  \frac { ( \mathrm { x } - 1 ) ^ { 2 } } { 36 } + \frac { ( \mathrm { y } + 1 ) ^ { 2 } } { 4 } = 1 .  Graph the new ellipse.

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Describe the graph of the polar equation. - r2sin2θ=22\mathrm { r } ^ { 2 } \sin 2 \theta = 22

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Find the foci of the ellipse. - 196x2+81y2=15,876196 \mathrm { x } ^ { 2 } + 81 \mathrm { y } ^ { 2 } = 15,876

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If the equation represents a hyperbola, find the center, foci, and asymptotes. If the equation represents an ellipse, find the center, vertices, and foci. If the equation represents a circle, find the center and radius. If the equation represents a parabola, find the focus and directrix. - 25x2+4y2300x40y+900=025 x ^ { 2 } + 4 y ^ { 2 } - 300 x - 40 y + 900 = 0

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Find the length of the curve. -The curve r=71+sin2θ,0θπ4r = 7 \sqrt { 1 + \sin 2 \theta } , 0 \leq \theta \leq \frac { \pi } { 4 }

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Graph the pair of parametric equations with the aid of a graphing calculator. - x=2cost+cos2t,y=2sintsin2t,0t2πx = 2 \cos t + \cos 2 t , y = 2 \sin t - \sin 2 t , 0 \leq t \leq 2 \pi  Graph the pair of parametric equations with the aid of a graphing calculator. - x = 2 \cos t + \cos 2 t , y = 2 \sin t - \sin 2 t , 0 \leq t \leq 2 \pi

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Solve the problem. -Find the volume generated by revolving about the yy -axis the area bounded by the curves 2x2y2=50,x=82 x ^ { 2 } - y ^ { 2 } = 50 , x = 8

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