Find the focus and directrix of the parabola x2 = 12y?
915081
the standard form of the equation for a parabola with vertex (0,0) is x^2 = 4py  where p is the distance from the vertex to the focus, which is equal to the distance from the vertex to the directrix.  So if x^2 = 12y, then 4p = 12, so p = 3.  This means the focus is 3 units in the positive direction from the vertex and the directrix is 3 units away in the negative direction.  Since the vertex is (0,0), the focus is (0,0+3)  or (0,3), and the directrix is the horizontal line y=-3.
893896
y=x^2\ny'=2x\n2x=tan60*\n=rt3\n=rt3/2\nsub y=(rt3/2)^2\n=3/4\nthe point is (rt3/2,3/4)
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
819984
restate the first equation in terms of y:\n\ny = 12 - x\n\nNow substitute this in for y in the second equation:\n\n2x² + 3(12 - x)² = 7x(12 - x)\n\nNow, multiply it out and solve for x:\n\n2x² + 3(144 - 24x + x²) = 84x - 7x²\n2x² + 3x² + 7x² - 72x - 84x + 432 = 0\n12x² - 156x + 432 = 0\n12(x² - 13x + 36) = 0\nx² - 13x + 36 = 0\n(x - 9)(x - 4) = 0\n\nx = 9 or 4\n\nSo, {x,y} = {9,3} or {4,8}
759663
a) The curve you need is called a cycloid. Look that up on the web.\n\nb) The parametric form of the cycloid is\n\nx=12(a-sin a)      y=12(1-cos a) where a is then angular displacement. 24pi (circumference) will be the period and it passes through (0,0)
804277
this is an example becuase im not doing it for you.\n\nThings you need to know\n\nFormula for a parabola (with vertical symmetry): (x-h)^2 = 4a(y-k) where "h" and "k" are the x and y corrdinates of the vertex, and "a" is the distance from the vertex to the focus.\n\nCoordinates of the vertex: (h,k) - these tell you how far to shift your parabola from the origin.\nCoordinates of the focus: (h, k+a) - if a is the distance from vertex to focus, then k+a should give you the point of the focus\nFormula of the directrix: y = k-a -- the vertex lies halfway between the focus and directrix, so subtracting the distance from the vertex to the focus will give you the directrix.\n\nPlease ask your teacher if you do not understand the reasoning behind any of these formulas.\n\n~~~~~~~~~~~~~~~~~~~~~\nNow, from what you are given, a vertex at (0,0) and a focus at (0,4) you can see that since the y coordinate is the one that changes, the focus lies above the vertex (not left or right) so it must be a vertically symmetrical parabola. You can also tell that it opens upward, since the focus lies inside the parabola and it is above the vertex. This means that the directrix is a flat line below the parabola.\n\n~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nOn to solving your problem:\n\nVertex (h,k) = (0,0)\nFocus (h,k+a) = (0,4) so k+a = 4 -- from the vertex you know that k=0, so 0+a = 4, or a = 4\n\nRemember the formula for the directrix was y = k-a. Plug in what you know and you get y = 0-4, or y = -4\n\nhope it help.... where's my 10 points?¿?
698387
y= -(2/3)x^2 + (2/3)x + 4\n\n\n(P): y = ax^2 + bx + c\n\nPoints (0;4)  (-2;0)  (3;0) are element of (P)\n-->  4 = 0a + 0b + c --> c = 4\n0 = 4a - 2b + c\n0 = 9a + 3b + c\n\n--> 5a + 5b = 0 --> b = -a\n--> 4a + 2a + 4 = 0 --> a = -2/3,  b=2/3
806655
dx=2xlny-2z\ndy=x^2*(1/y)-2z\ndz= x^2lny-2
1118234
first you have to put the hyperbola in standard form.  Divide everything by 4.\n\n9x^2 - y^2 = 1\nWhich is the same as\n(x^2)/(1/9) - (y^2)/1 = 1\nHere, a^2 = 1/9, so a = 1/3, \nand b^2 = 1, so b = 1\n\nIn this case, the center is (0,0)\nSince the "x" term is first, then it opens left and right.  The vertices will be left and right from the center by "a" units, so the vertices are (1/3, 0) and (-1/3, 0).\n\nThe foci are left and right of the center by "c" units.  In a hyperbola, a^2 + b^2 = c^2, so c^2 = (1/3)^2 + (1^2) = 1/9 + 1 = 10/9.  c = root(10)/3.  The foci will be (root10/3, 0) and (-root10/3, 0).\n\nThe asymptotes have slopes that are +b/a and -b/a, and pass through the center, in this case, (0,0).  So these slopes are +1/(1/3) and -1/(1/3) or 3 and -3, and the equations are y = 3x and y = -3x.
934203
y + 2y - 10 = 2y + 2\n3y - 10 = 2y + 2\n3y = 2y + 12\ny = 12
1037602
A parabola is the set of all points equidistant from a fixed point and a fixed line in the plane.  The fixed line is the directrix.  Since you are wanting a directrix y = -4, you are wanting a parabola that opens upward.  \n\nThe general form for a parabola that opens upward is:\n\n(x-h)^2 = 4p(y-k)\n\nwhere (h,k) is the vertex and y = -p is the directrix.\n\nSince you want a vertex of (0, 0), this means that your h and k are both 0.  The p in question is always greater than zero.  You have y = -4, so p = 4.  \n\nTherefore, we have (x-0)^2 = 4*4*(y-0)\n\nSo our equation is x^2 = 16y.\n\nAs far as the the standard form goes, different books refer to various forms as "standard."  The most common standard form for a parabola of this nature is to solve for y.\n\nSo we get:  y = (1/16)x^2\n\nAngel
1079894
The basic definition of a parabola states that it is the locus of points that are equidistant from a given point (the focus), and a line (the directrix). Therefore, we have to find two distances:\n\nPoint-to-Point: d1 = sqrt( (x-11)^2 + (y+7)^2 )\nPoint-to-Line: d2 = abs( x-5 ) / sqrt (1^2) = (x - 5)\n\nThen, make the two distances equal, d1 = d2:\n(x-11)^2 + (y+7)^2 = (x-5)^2\n........simplifying........\ny^2 + 14y + 49 - 12x + 96 = 0\n\n(y + 7)^2 = 12*(x - 8)\n\nThis is the equation for our parabola.
804813
(y+1)(y-7)\n=y^2-6y-7
1117529
y = 3x^2 + 12x - 7\nWhat do the directions say?
583382
Try the moebius spiral. It's pretty interesting because it's non orientable. \n\nIt's 3d (2 parameters s and t) and described by the parametrized vector components:\n\n[3*cos(s)+t*sin(s/2),\n3*sin(s),\nt*cos(s/2)]\n\n(every component of the vector in a new line)\n\nThe parameter ranges are:\n\ns=-2*Pi...2*Pi,t=-1..1\n\nMAPLE INSTRUCTIONS TO PLOT IT:\n\nwith(plots);\nplot3d([3*cos(s)+t*sin(s/2),3*sin(s),t*cos(s/2)],s=-2*Pi...2*Pi,t=-1..1);\n\nSorry, that instruction is big and not displayed well, but I guess since you know the parametrized vector you can find it.\n\nWHAT DO YOU WANT IT FOR? WANT TO PUBLISH ANYTHING?
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30
979245
this isa parabola opening downwards\nvertices are 1 and 3\na=-2\ngive x different values say -3,-2,-1,0,1,2,3 and fing the corresponding values of y\ntabulate and plot the ordered pairs