Lesson 3 (Mulit. Choice Questions)

Halley’s comet has an elliptical orbit with the sun at one focus. Its orbit shown below is given approximately by In the formula, r is measured in astronomical units. (One astronomical unit is the average distance from Earth to the sun, approximately 93 million miles.) Find the distance from Halley’s comet to the sun at its greatest distance from the sun. Round to the nearest hundredth of an astronomical unit and the nearest million miles.

	A. 12.13 astronomical units; 1128 million miles
	B. 91.54 astronomical units; 8513 million miles
	C. 5.69 astronomical units; 529 million miles
	D. 6.06 astronomical units; 564 million miles

Use the center, vertices, and asymptotes to graph the hyperbola.

(x – 1)² – 9(y – 2)²= 9

	A.
	B.
	C.
	D.

Find the standard form of the equation of the ellipse and give the location of its foci.

	A. + = 1 foci at (- , 0) and ( , 0)
	B. – = 1 foci at (- , 0) and ( , 0)
	C. + = 1 foci at (- , 0) and ( , 0)
	D. + = 1 foci at (-7, 0) and ( 7, 0)

Rewrite the equation in a rotated x’y’-system without an x’y’ term. Express the equation involving x’ and y’ in the standard form of a conic section.

31x² + 10xy + 21y²-144 = 0

	A. x‘2 = -4 y’
	B. y‘2 = -4x’
	C. + = 1
	D. + = 1

Find the standard form of the equation of the ellipse satisfying the given conditions. Foci: (0, -2), (0, 2); y-intercepts: -5 and 5

	A. + = 1
	B. + = 1
	C. + = 1
	D. + = 1

Find the vertices and locate the foci for the hyperbola whose equation is given.

49x² – 100y²= 4900

	A. vertices: ( -10, 0), ( 10, 0) foci: (- , 0), ( , 0)
	B. vertices: ( -10, 0), ( 10, 0) foci: (- , 0), ( , 0)
	C. vertices: ( -7, 0), ( 7, 0) foci: (- , 0), ( , 0)
	D. vertices: (0, -10), (0, 10) foci: (0, – ), (0, )

Write the equation in terms of a rotated x’y’-system using θ, the angle of rotation. Write the equation involving x’ and y’ in standard form. xy +16 = 0; θ = 45°

	A. +  = 1
	B. y‘2 = -32x’
	C. + = 1
	D. – = 1

Write the appropriate rotation formulas so that in a rotated system the equation has no x’y’-term.

10x² – 4xy + 6y²– 8x + 8y = 0

	A. x = -y’; y = x’
	B. x = x’ – y’; y = x’ + y’
	C. x = (x’ – y’); y = (x’ + y’)
	D. x = x’ – y’; y = x’ + y’

Find the location of the center, vertices, and foci for the hyperbola described by the equation.

– = 1

	A. Center: ( -4, 1); Vertices: ( -10, 1) and ( 2, 1); Foci: and (
	B. Center: ( -4, 1); Vertices: ( -9, 1) and ( 3, 1); Foci: ( -3 + , 2) and ( 2 + , 2)
	C. Center: ( -4, 1); Vertices: ( -10, -1) and ( 2, -1); Foci: ( -4 – , -1) and ( -4 + , -1)
	D. Center: ( 4, -1); Vertices: ( -2, -1) and ( 10, -1); Foci: and

Sketch the plane curve represented by the given parametric equations. Then use interval notation to give the relation’s domain and range.

x = 2t, y = t²+ t + 3

	A. Domain: (-∞, ∞); Range: -1x, ∞)
	B. Domain: (-∞, ∞); Range: [ 2.75, ∞)
	C. Domain: (-∞, ∞); Range: [ 3, ∞)
	D. Domain: (-∞, ∞); Range: [ 2.75, ∞)

Use vertices and asymptotes to graph the hyperbola. Find the equations of the asymptotes.

y = ±

	A. Asymptotes: y = ± x
	B. Asymptotes: y = ± x
	C. Asymptotes: y = ± x
	D. Asymptotes: y = ± x

Graph the ellipse.

16(x – 1)² + 9(y + 2)²= 144

	A.
	B.
	C.
	D.

Is the relation a function?

y = x²+ 12x + 31

	A. Yes
	B. No

Determine the direction in which the parabola opens, and the vertex.

y²= + 6x + 14

	A. Opens upward; ( -3, 5)
	B. Opens upward; ( 3, 5)
	C. Opens to the right; ( 5, 3)
	D. Opens to the right; ( 5, -3)

Match the equation to the graph.

x²= 7y

	A.
	B.
	C.
	D.

y²= -2x

	A.
	B.
	C.
	D.

Convert the equation to the standard form for a hyperbola by completing the square on x and y.

x² – y²+ 6x – 4y + 4 = 0

	A. (x + 3)2 + (y + 2)2 = 1
	B. – = 1
	C. (x + 3)2 – (y + 2)2 = 1
	D. (y + 3)2– (x + 2)2 = 1

Eliminate the parameter t. Find a rectangular equation for the plane curve defined by the parametric equations.

x = 6 cos t, y = 6 sin t; 0 ≤ t ≤ 2π

	A. x2 – y2 = 6; -6 ≤ x ≤ 6
	B. x2 – y2 = 36; -6 ≤ x ≤ 6
	C. x2 + y2 = 6; -6 ≤ x ≤ 6
	D. x2 + y2 = 36; -6 ≤ x ≤ 6

Convert the equation to the standard form for a parabola by completing the square on x or y as appropriate.

y²+ 2y – 2x – 3 = 0

	A. (y + 1)2 = 2(x + 2)
	B. (y – 1)2 = -2(x + 2)
	C. (y + 1)2 = 2(x – 2)
	D. (y – 1)2 = 2(x + 2)

Convert the equation to the standard form for a hyperbola by completing the square on x and y.

y² – 25x²+ 4y + 50x – 46 = 0

	A. – (x – 2)2 = 1
	B. – (y – 1)2 = 1
	C. (x – 1)2 – = 1
	D. – (x – 1)2 = 1