Find the standard form of the equation of the parabola with focus (−10, 0) and directrix x=10.
y2=−10x
y2=−40x
x2=−40y
y2=40x
Convert the equation y2−4y−5x+24=0 to the standard form for a parabola by completing the square.
(y+2)2=5(x−4)
(y−2)2=5(x−4)
(y+2)2=−5(x−4)
(y−2)2=5(x+4)
Find the focus and directrix of the parabola with equation x=7y2.
focus: (128, 0); directrix: x=−128
focus: (0,128); directrix: y=−128
focus: (128, 0); directrix: x=128
focus: (17, 0) directrix: x=−17
Graph the parabola with equation y2=−7x.
Find the vertex, focus, and directrix of the parabola with equation (x−3)2=12(y−1).
vertex: (3, 1); focus: (3, 4); directrix: y=−2
vertex: (3, −1); focus: (−3, 2); directrix: y=−4
vertex: (3, 1); focus: (3, −2); directrix: x=4
vertex: (1, 3); focus: (1, 6); directrix: y=0
The parabolic arch of a bridge has a 180-foot base and a height of 25 feet. Find the height of the arch 45 feet from the center of the base.
12.5 ft
6.25 ft
16.7 ft
18.75 ft
Find an equation of the parabola with vertex (−2, 1) and directrix x=2.
(y−1)2=−16(x+2)
(y−1)2=16(x+2)
(y+2)2=−16(x−1)
(y+2)2=16(x−1)
In Problems 8–10, find the standard form of the equation of the ellipse satisfying the given conditions.
Foci: (−3, 0), (3, 0); vertices: (−5, 0), (5, 0)
x29+y216=1
x225+y216=1
x29+y225=1
x216+y225=1
Foci: (0, −3), (0, 3); y-intercepts: −7, 7
x249+y240=1
x29+y249=1
x29+y240=1
x240+y249=1
Major axis vertical with length 16; length of minor axis=8; center: (0, 0)
x216+y264=1
x264+y2256=1
x28+y264=1
x264+y216=1
Graph the ellipse with equation 9(x−1)2+4(y−2)2=36.
Find the vertices and foci of the hyperbola with equation x2121−y24=1.
vertices: (−11, 0), (11, 0); foci: (−2, 0), (2, 0)
vertices: (0, −1), (0, 11); foci: (−55–√, 0)(55–√, 0)
vertices: (−11, 0), (11, 0); foci: (−55–√, 0)(55–√, 0)
vertices: (−2, 0), (2, 0); foci: (−55–√, 0)(55–√, 0)
Graph the hyperbola with equation x225−y24=1.
Find the standard form of the equation of the hyperbola with foci (0, −10), (0, 10) and vertices (0, −5), (0, 5).
y225−x2100=1
y225−x275=1
x225−y2100=1
x225−y275=1
In Problems 15 and 16, convert the equation to the standard form for a hyperbola by completing the square on x and y.
y2−4x2−2y−16x−19=0
(y−2)24−(x+4)2=1
(x−1)24−(y+2)2=1
(y−1)24−(x+2)2=1
(x+2)2−(y−1)24=1
4y2−9x2−16y−36x−56=0
(y−2)24−(x+2)29=1
(x−2)24−(y+2)29=1
(y−2)29−(x+2)24=1
(y+2)29−(x−2)24=1
In Problems 17–20, identify the conic section.
3x2+2y2−6x+4y−1=0
parabola
circle
ellipse
hyperbola
x2+2x+4y+5=0
parabola
circle
ellipse
hyperbola
4x2−y2+16x+2y+11=0
parabola
circle
ellipse
hyperbola
5y2−3x2+20y−6x+2=0
parabola
circle
ellipse
hyperbola
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