10.6 Parametric Forms Of Equations
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Identify the conic x2-4y-6x+9=0
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Circle
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Ellipse
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Hyperbola
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Parabola
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- 2.
Write in standard form x2-4y-6x+9=0
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4y=(x-3)^2
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4y=(x+3)^2
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4y=(x+2)^2
Correct Answer
A. 4y=(x-3)^2Explanation
The given equation can be rearranged to the standard form of a quadratic equation, which is Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. In this case, the equation is x^2 - 4y - 6x + 9 = 0. By simplifying the equation, we get 4y = (x - 3)^2. Therefore, the correct answer is 4y = (x - 3)^2.Rate this question:
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- 3.
X2- 3y2 +2x - 24y -41 = 0identify the conic
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Ellipse
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Parabola
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Hyperbola
Correct Answer
A. HyperbolaExplanation
The equation x^2 - 3y^2 + 2x - 24y - 41 = 0 is a hyperbola because it is a second-degree equation with both x^2 and y^2 terms, and the coefficients of x^2 and y^2 have opposite signs. This indicates that the graph of the equation will be a hyperbola.Rate this question:
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- 4.
X2- 3y2 +2x - 24y -41 = 0write in standard form
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(x+4)^2 / 2 - (y+1)^2 / 6 = 1
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(y+4)^2 / 2 - (x+1)^2 / 6 = 1
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(x-4)^2 / 2 - (y+1)^2 / 6 = 1
Correct Answer
A. (y+4)^2 / 2 - (x+1)^2 / 6 = 1Explanation
The given equation is in the standard form of a hyperbola, which is (y+4)^2 / 2 - (x+1)^2 / 6 = 1. This is determined by comparing the given equation with the general equation of a hyperbola, (x-h)^2 / a^2 - (y-k)^2 / b^2 = 1, where (h,k) represents the center of the hyperbola and a and b are the distances from the center to the vertices. In this case, the center is (-1,-4), a^2 = 6, and b^2 = 2. Therefore, the correct answer is (y+4)^2 / 2 - (x+1)^2 / 6 = 1.Rate this question:
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- 5.
Find the rectangular equationx = t, y = 2t2 - 4t + 1 ; -infinity
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Y = 2x^2 +4x +1
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Y = 2x^2 - 4x +1
Correct Answer
A. Y = 2x^2 - 4x +1Explanation
The given equations are parametric equations in terms of the parameter t. To convert them into rectangular equations, we can substitute x = t into the equation for y. By doing so, we get y = 2(t)^2 - 4(t) + 1, which simplifies to y = 2t^2 - 4t + 1. Therefore, the correct rectangular equation is y = 2x^2 - 4x + 1.Rate this question:
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- 6.
Find the rectangular equationx = cos 2t, y=sin 2t; 0<=t<=2pi
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X^2 + y^2 =1
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X^2 + 2y^2 = 1
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2x^2 + 2y^2 = 1
Correct Answer
A. X^2 + y^2 =1Explanation
The given parametric equations x = cos 2t and y = sin 2t represent a parametric curve in the xy-plane. By substituting these equations into the equation x^2 + y^2 = 1, we can verify that it holds true for all values of t between 0 and 2pi. This equation represents a unit circle centered at the origin, which means that any point (x, y) on the curve satisfies the equation x^2 + y^2 = 1. Therefore, the correct answer is x^2 + y^2 = 1.Rate this question:
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- 7.
X = -cos t, y = sin t, 0<=t<=2pi the rectangular equation for the given problem is the same which of the following:
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X=3sint, y=2cost
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X=cos 2t, y=sin 2t
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X=1, y=1
Correct Answer
A. X=cos 2t, y=sin 2tExplanation
The given parametric equations x = -cos t and y = sin t represent a circle centered at the origin with radius 1. In the answer choice x = cos 2t and y = sin 2t, the parametric equations represent the same circle, but with the parameter t multiplied by 2. Therefore, the rectangular equation for the given problem is x = cos 2t and y = sin 2t.Rate this question:
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- 8.
In the general form of a conic, we can guess that it is a parabola if and only if x is squared and the y term is not squared
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True
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False
Correct Answer
A. FalseExplanation
The explanation for the given correct answer is that the statement is incorrect. In the general form of a conic, a parabola can be identified if and only if the x term is squared and the y term is not squared. Therefore, the correct answer is false, as it contradicts the conditions for identifying a parabola.Rate this question:
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- 9.
Identify the conic represented by the rectangular form of the equation:x= - cos t, y = sin2t, 0<=t<=180
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Ellipse
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Parabola
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Hyperbola
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Circle
Correct Answer
A. ParabolaExplanation
The equation x = -cos(t) represents a horizontal line that moves from -1 to 1 as t varies from 0 to 180. The equation y = sin^2(t) represents a curve that oscillates between 0 and 1 as t varies from 0 to 180. When we plot these points on a graph, we can see that they form a parabolic shape. Therefore, the conic represented by the given equation is a parabola.Rate this question:
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- 10.
After drawing a parametric graph, we need to:
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Shade it
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Draw arrows using the t values
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Identify the vertex
Correct Answer
A. Draw arrows using the t valuesExplanation
After drawing a parametric graph, we need to draw arrows using the t values. This is because the t values represent the parameter that determines the position of the point on the graph. By drawing arrows along the graph using the t values, we can indicate the direction in which the parameter is increasing or decreasing. This helps in understanding the movement and behavior of the graph.Rate this question:
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Dec 11, 2023Quiz Edited by
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May 01, 2010Quiz Created by
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