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Sample Exam 1 Fall 2011

# Sample Exam 1 Fall 2011 - Math 2123.2— Fall 2011 —...

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Unformatted text preview: Math 2123.2— Fall 2011 — Sample Exam 1 Solution To receive credit, you must show your work. In all graphs, the marks on the axes are one unit apart. 1. [15 points] Plot the points (—1,—2) and (—2, 3) on the Cartesian coordinate system shown below. Also determine the distance between these points. The distance between these points is (—1 ﬂ —2)2 + (—2 — 3)2 = \/12 + (—5)? : fig—6 m 5.099 2. The ﬁgure below shows a line in a Cartesian coordinate system. Answer the following questions about the line. (a) [5 points] What is the slope of the line? (b) [5 points] What is the y-intercept of the line? ) (c [3 points] What is the equation of the line in slope-intercept form? (a) The line falls one unit for three units across, so its slope is —1/3. (These distances are shown on the diagram in green.) (b) The y—intercept is (0,1). (You could also just say that it is 1. The relevant point is circled in red on the diagram.) (c) The equation of the line in slope—intercept form is 1 = ——:2: + 1. y 3 3. A circle has center at (~1, 2) and passes through the point (3, 3). (a) [10 points] Write the equation of the circle in standard form. (b) [5 points] Does the point (2, —1) lie on this circle? Why or why not? (a) The equation must be (\$ ~ —1)2 + (y — 2)2 (x + l)2 + (y — 2)2 : some constant. some constant To ﬁnd the constant, we use the fact that (3, 3) is on the circle, so putting a: = 3 and y = 3 makes the equation true: (3+1)2+(3—2)2=42+12:17 Thus the ﬁnal equation is (:0 + 1)2 + (y — 2)2 = 17. (b) We check by substituting a: : 2 and y = —1 into the equation: (2 + 1)2 + (—1 — 2)2 = 17 This simpliﬁes to 18 = 17, which is false. The point does not lie on the circle. 4. Consider the circle with equation x2+y2+4x—6y+11:0. (a) [8 points] Find the center of the circle. (b) [7 points] Find the radius of the circle. (c) [5 points] Sketch the circle on the Cartesian coordinate system shown below. x2+y2+4x—6y+1120 [x2+4\$]+[y2—6y]+11 =0 [(x+2)2 4] 1 [(y 3)2 9] 111— 0 WWWPWEW»? WW (x -l- 2)2 -l- (y — 3)2 — 2 = 0 (m+2)2+(y—3)2=2 (a) The center of the circle is (—2, 3). (b) The radius of the circle is M? m 1.414. 5. [12 points] The line L is perpendicular to the line through the points (1, 2) and (3, 5). What is the slope of L? The line through the points (1, 2) and (3, 5) has slope 5—2_§ 3—1—2' The slope of the perpendicular line L is the negative reciprocal of this, which is —2 / 3. 6. A parabola has vertex at (O, 0), axis the x-axis, and passes through the point G, 2). a) 8 points] Find the equation of the parabola. ( w l ) [5 points] Find the focus of the parabola. (c) [5 points] Find the directrix of the parabola. (d) [7 points] Sketch the parabola, the directrix, and the focus on the Cartesian coordinate system below. (a) Since the axis is the x—axis and the ver— tex is at (0,0), the equation is y2 2 4pm for some constant p. The point (1/2, 2) is on the parabola and so this point makes the equation true. That is, 22 = 4p(1/2) or p = 2. The equation is y2 = 835. (b) Since p 2 2 and the axis is the x-axis, the focus is (2, 0). (It is shown as a cross in the ﬁgure.) (0) The directrix is m = —2. (It is shown as a dashed line in the ﬁgure.) Math 2123.2 — Fall 2011 — Some Alternate Problems for Sample Exam 1 Solutions 1. [13 points] Find the equation of the line that passes through the midpoint of the line segment from (—1, 3) to (3, —5) and is perpendicular to this segment. —1 3 3 —5 . , The midpoint of the segment is < 2+ , +2 > = (1, — 1). The slope of the segment is 5 3 8 = — : —2. 3 — —1 4 The slope of the perpendicular line is the negative reciprocal, which is 1/2. By the point—slope form, the equation of the line is + 1 1 1 = —\$ — — y 2 2 _ 13: 3 y _ 2 2' 2. Consider the ellipse with equation 43:2 + 3/2 = 4. (a) [8 points] Find the vertices of the ellipse. (b) [5 points] Find the foci of the ellipse. ) (c [7 points] Sketch the ellipse, the vertices, and the foci on the Cartesian coordinate system shown below. 49? + y2 = 4 \$2 y2 F+¥Zl (a) The vertices are (0, i2). WWW (b) The distance from the center at (0, 0) to the foci is «22 — 12 = \/E m 1.732. The y-axis is the major axis and so the foci are (0, :\/§). 3. Consider the hyperbola with equation 9342 —a:2 = 9. (a) [5 points] Find the vertices of the hyperbola. (b) [5 points] Find the asymptotes of the hyperbola. (C) [10 points] Sketch the auxiliary rectangle, the asyniptotes, and the hyperbola on the Cartesian coordinate system shown below. 9y2~cc2=9 2 2 y x 15_§:1 (a) The vertices are at (O, :1). (b) The asymptotes are \$ =i—. y 3 Tghis can be seen by solving the equation y— — £3 — O for y. The corners of the auxiliary rectangle are at (3:3, :lzl). The asymptotes pass through opposite corners of this rectangle. The foci were not actually requested in this prob-- lem, but I put them in anyway. They are at (0, ::\/ 10) and are indicated by crosses. ...
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