University:
Florida Gateway College
Course:
MAT 0028 | Elementary Algebra
Academic year:

2019
Views:

134

Pages:

5
Author:

elestallidofm6q

9.1-9.3 Quiz REVIEW HPC Name: Keef Hour: Find the standard form of the equation of the circle with the given characteristics. 1. Center: (0,0); point on circle: (8,-15) 2. Endpoints of diameter: (-2,3) and (6,-5) (x-4)²+(y-k)²=r Center: 22 (8-0)+(-15-0)=2 (x-4)+(y-k) 8+(-15)² (-2-2)+(3-1-1)) 64+225c² (4)+(4) (x-2)+(y+1)=32 289 x²+y²=289 32= Write the equation of the circle in standard form. Sketch the circle and then identify its center, radius, x-intercepts, and y-intercepts. 3. x²+y² +4x+6y-3=0 x²+4x+4 +y²+by+9 = 3+4+9 (x+2)+(y+3)16 Center: (-2,-3) Radius: 4 x-intercepts: y-intercepts: (x+2)²+(0+3)16 (0+2)²+(y+3)=16 4+(y+3)=16 (x+2)+9016 (+5)=12 (x+2)=7 Y+3=±213 メー Y=-3125 4. 4x²+4y² +32x-24y +51 = 0 4x²+32x+4y²-24y = -51 4/x²+8x46)+4(4-6419)=-5464+36 4(x+4)²+4(y-3)= 49 (x+4)²+(y-3)²= 4 Center: (-4,3) Radius: x-intercepts: y-intercepts: Nowe! Can ses from graph, but cheche algebraically (0+4)+(y-3)= 14+(y-3)²= 4 N (x+4)=± - focus, and directrix. 5. y²-4y4x= 0 Y²-4y+4= 4x +4 (4-2)²=4(x+1) Vertex: (-1, 2) 4p=4 P1 opens right Focus: (0,2) Directrix: 2 6. x²+4x+6y-2=0 x²+4x+4= -by+2+4 (x+2)=-64+6 40=-6 (x+2)²=-6(y-1) P=-3/2 gens down Vertex: (-2,1) Focus: (-2,-) Directrix: Y= Find the standard form of the equation of the parabola with the given characteristics. 8. Focus: (8,-2): Directrix: x=4 7. Vertex: (4, 2); Focus: (4, 0) (-k)=4ply-k) (x-4)=4(-2xy-2) ogens down (x-4)=-8(y-2) P=-2 Vertex = (6-2) (-k)=4p(x-h) (y+2)=4(2)(x-4) (y+2): 8(x-6) N N geens right y 9 . A satellite dish has the shape of a paraboloid. The signals that it receives are reflected to a receiver that is located at the focus of the paraboloid. If the dish is 8 feet across at its opening and 1.25 feet deep at its center, determine the location of its focus. B (0,0) 41.25 (x-4)=4p(y-k) (4-0) 4p(1.25-0) 16 = Sp pollo/g 3.2 feet 10. A parabolic archway is 12 meters high at its vertex. At a height of 10 meters, the width of the archway is 8 meters. How wide is the archway at ground level? (4,10) (9.8,0) (0,12) ( x-1)²=4ply-k) (4-0): 4p(10-12) x=-8(-12) X-8(0-12) 16-80 X 96 (98,0) P-2 X= 9.8 ground lovel goud level 19.6 m 2 Write the equation of the ellipse in standard form. Sketch the ellipse and then identify its center, vertices, co-vertices, foci and eccentricity. 11. 4x²+25y² + 16x150y+1410 4x²+16x 1254-1508-141 4(x+2)²+25(y-3): 100 4(x²+4x+1) +25(y-by+9)= -141+the+22s 4 (x+2)+(y-3) 25 a=5 6-2 Center: (-2, 3) Vertices: (-7,3) (-2,1) (3,3) Co-Vertices: (-2,5) Foci: ab C= 2 Eccentricity: 25-4 e: 21 a 5 (-2±√21,3) 12. 16x+9y32x+72y +16=0 lok 32x +9y+72y-16 16(x²-2x+1) +9(y²+8y+16)=-16+16+144 16(x-1)² +9 (y+4)=144 (x-1)+(y+4)=1 9 6:3 مال Center: (1,-4) Vertices: (1,0) Co-Vertices: (4,-4) (-2,4) Foci: ab Eccentricity: a 16-9 (1,-4±√7) (18) 13. Find the standard form of the equation of the ellipse with the following characteristics: 6=5 Center at (4.-2), Foci at (4.1) and (4,-5), and minor axis of length 10. 0:3 (x-4)²+(y+2)²=1 25 34 9-02-25 14. A sunburst window above a doorway is constructed in the shape of a top half of a horizontally oriented ellipse. The window is 20 in, tall at its highest point, and 80 in. wide at the bottom. Find the height of the window, 25 in. from the center of the base. 20 (25,y) x-h)+(y-k) ( 42 2 1600 400 40 25 1600+400=1 ५२ 1 = 400 1000 2 625+4=1 1600 400 390000-1140 Y2243.75 Y=15.6m

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