how can i write an equation of a parabola with the ends of the latus rectum at (2,-2)&(2,8)&the vertex @ (7,3)
412752
This looks like it's been rotated 90 degrees counterclockwise.\n\nSo, I guess it would be as follows:\n\nx = -(9/25)(y-3)^2 + 7
1185644
u got it wrong\nx=at^2 and y=2at\n\nparabola y^2=4ax\n=> y^2 / 4a = x\n=> y^2 / 4a^2 = x/a\nlet this b equal to t^2\n\nthus \ny^2 / 4 a^2 = x/ a = t^2\nthus y= 2at and x=at^2
953526
Parabola\nThe vertex of the parabola used to be at the origin. The vertex will now be at the point (h,k). \n\nVertical Axis of Symmetry\n( x - h )2 = 4 p ( y - k )\n\nThe vertex is at (h,k). \n\nSince the axis of symmetry is vertical, the foci and directrix have different y coordinates than the vertex. \n\nSince p is the directed distance from the vertex to the focus, the focus will be at (h,k+p) \n\nSince the directrix is p units the other side of the vertex, the equation of the directrix will be y=k-p\n\nThe parabola will open up if p is positive and down if p is negative.
903023
M (-7, -3) is the midpoint of segment PQ and the coordinates of Q are (6, -3).\nLet the co-ordinates of P be (x,y).\nThen, (-7,-3) = ((x+6)/2, (y-3)/2)\n=>(x+6)/2 => -7 , (y-3)/2 = -3\n=>(x+6) => -14 , (y-3) = -6\n=>x => -14-6 , y = -6+3\n=>x => -20 , y = -3\n\nSo, co-ordinates of point p are (-20, -3)\n\nAll the best
741586
set y=0\nsolve quadratic eqn\nquadratic eqn\n\n-7 +- (49-16)^.5\n-------------------\n2\n= (-7 +- 33^.5)/2
1050977
http://mscir.tripod.com/parabola/
883980
find the mid pointof any side by using the formula (x1+x2/2,y1+y2/2)\nusing that and the opposite vertex using the two point formula (y-y1/y2-y1)=(x-x1/x2-x1) find the equation\ni shall do one for you\nbid point of BC=(0,-2)\npoint A=(0,7)\naquation of median AD=(y+2)/(7+2)=(x-0)/(0-0)\n0(y+2)=9(x-0)\n9x=0 is th equation of median AD
434174
If the equation is y = 4x^2 + 4x - 5, then:\n\nRewrite the equation and complete the sqaure on the right side of the equation...\n\ny = 4(x^2 + x) - 5\n\ny = 4(x^2 + x + 1/4) - 5 -4(1/4)\n\ny = 4(x + 1/2)^2 - 5 - 1\n\ny = 4(x + 1/2)^2 - 6\n\ny - (-6) = 4(x + 1/2)^2\n\ny - (-6) = 4[x - (-1/2)]^2\n\nVertex for standard equation is in the form:\n\ny - k = +/- a(x - h)^2\nwhere:\nh = x-coordinate of the vertex.\nk = y-coordinate of the vertex.\n+ a = the parabola opens upward.\n- a = the parabola opens downward\na = the parabola will have a wide opening if a > 1.\na = the parabola will have a narrow opening if a < 1.\n\nTherefore vertex is at (-1/2, -6) and a wider opening which is upward.
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.
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?¿?
699042
Get your formula into this format:\n\ny = (x+a)²\n\nWhen it's in this format, it's easy to get the axis of symmetry. The axis of symmetry is the line x = -a.  (This is also the value (x=-a) that will give you the vertex of your parabola.  The vertex and line of symmetry are related, because the vertex always goes through the line of symmetry.)\n\nSo since you have y = (x+2) ² your axis of symmetry is the line x = -2.\n\nHope that helped!\n\n~ ♥ ~
699742
-(3x+y)(3x-y)\n-(9x^2-y^2)\ny^2-9x^2
736379
for this just substitute for x and y in eqn and see if LHS=RHS\nLHS=4\nRHS=(-1/2)(-3)-5/2\n=3/2-5/2\n=-2/2\n=-1\nnot = RHS\nso pt. not on line
1147966
Yes!!!!!!!!! Email me, and I'll try to shed more light on this. Parabolas are where my genius comes out :)
670280
If you have a parabola in the form of y = a(x - h)^2 + k, then you can read off a lot of your particulars directly from the equation:\n\nvertex: (h, k)\ndirection: for a>0, parabola opens up, for a<0, parabola opens down\naxis of symmetry: x = h\n\nSo you have a parabola that opens down with vertex at (-2, 3), and axis of symmetry at x = -2.\n\nNow compare this to y = x^2: opens up, vertex is at (0, 0), and axis of symmetry at y = 0.  Your first equation will also be skinnier because of the a value: if |a|<1, your parabola is wider than y = x^2, and if |a|>1, then your parabola is skinnier than y = x^2.
450918
I think your equation should say “h = -16t^2 + 32t” - the “-162^2” part doesn’t make sense.  He goes up and down following a parabola, so there has to be a t^2 term somewhere for this equation to work.\n\nI’m going to assume that it should be -16t^2 and go from there.  To find the point where he gets to 14 feet, simply set the equation to 14:\n\nh = -16t^2 + 32t\n14 = -16t^2 + 32t\n0 = -16t^2 + 32t – 14\n0 = 8t^2 - 16t + 7\nt = [16 +/- sqrt((-16)^2 - 4(8)(7))] / 2(8)\nt = [16 +/- sqrt(256 - 224)] / 16\nt = [16 +/- sqrt(32)] / 16\nt = 1 +/- sqrt(2)/4\nt = 0.65s, 1.35s\n\nSo at 0.65s and 1.35s, he was at 14 feet.  Subtract the two and you find that he was at or above 14 feet for 0.7s.  (For a precise time, you can subtract [1 + sqrt(2)/4] - [1 - sqrt(2)/4] and get sqrt(2)/2.)
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)