Find an equation of the hyperbola?
960038
Since the foci of this hyperbola both lie on the horizontal line y=7, then the major axis is horizontal and the minor axis is vertical, and so the equation for it is \n\n((x-a)^2)/(M^2) - ((y-b)^2)/(m^2) = 1 \n\nwhere the center of the hyperbola (the midpoint of the segment between the two foci) is (a,b), M is the length of the semi-major axis (half the distance between the vertices),  and m is the length of the semi-minor axis.  If d is half the distance between the foci, then M^2 + m^2 = d^2\nThe center of this hyperbola is (7,2) since it is the midpoint of the segment between the foci.  M = 3 (the distance from the center to each vertex), d = 5 (the distance from the center to each focus), and so m = 4 since 3^2 + 4^2 = 5^2.  \n\nThus this hyperbola has equation \n\n((x-7)^2)/9 - ((y-2)^2)/16 = 1
296452
for a parabola, y = ax^2 + bx + c, so plug in the x,y coordinates of the points:\n\nP:  3 = a(7^2) + b(7) + c\nQ:  7 = a(9^2) + b(9) + c\nR:  1 = a((-2)^2) + b(-2) + c\n\nyou get:\n49a + 7b + c = 3\n81a + 9b + c = 7\n4a - 2b + c = 1\n\nsubtract equation 2 from eq 1 and eq 3 from eq 1 to get\n32a + 2b = 4\n45a + 9b = 2\n\nnow it gets ugly.  b has smallest coefficients, so divide eq 1 by 2 and eq 2 by 9 to get\n16a + b = 2\n5a + b = 2/9\n\nsubtract to get 11a = 16/9, then divide by 11:\na = 16/99.  plug this into the easiest equation, 5a + b... \n\n5(16/99) + b = 2/9\n80/99 + b = 2/9 = 22/99\nb = -58/99\n\nthen plug both a and b into an earlier equation:\n4a - 2b + c = 1\n4(16/99) - 2(-58/99) + c = 1\n64/99 + 116/99 + c = 1\n180/99 + c = 1\nc = 99/99 - 180/99 = 81/99 = 9/11\n\nso the parabola is y = 16/99 x^2 - 58/99 x + 9/11\n\nwasn't that fun?
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.
538371
http://en.wikipedia.org/wiki/Cubic_equation
889253
h^2 = p^2+b^2\n\nwhere h--- hypotenuse\n          p--- perpendicular\n          b--- base
554946
Use the equation d = sqrt((y2-y1)^2 + (x2-x1)^2)\nyou get d = sqrt((5-2)^2 + (3-(-1))^2)\n=> d = sqrt(9 + 16)\n=> d = 5 units
33948
u know general equation of circle is\n(x-h)^2+(y-k)^2=r^2\nwhere h and k are x and y coordinates of center of circle and r is radius of circle.\njust put the known values of h,k and r to get the equation.\nhey,i did it.
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~ ♥ ~
133564
The coordinates of the vertex of a parabola \ny = ax^2 + bx + c is\n[-b/2a , (4ac - b^2)/4a]\nJust substitute and you will get\nV(2 , -7)
483089
First write your equation using hyperbolic function:\n\ny = a cosh(x/a)\n\nDifferentiate, dy/dx = sinh(x/a)\n1 + (dy/dx)^2 = 1 + sinh^2(x/a) = cosh^2(x/a)\nSQRT (1 + (dy/dx)^2 ) = cosh(x/a)\n\nIntegrate from x = 0 to x = a:\n\nAnswer = [asinh(x/a)]@x=a  - [asinh(x/a)]@x=0\n       = asinh1 - asinh0\n       = asinh1\n       = a(e - (1/e))/2\n\nRefer to the following website for details on the catenary
187890
This question does not require the sledgehammer of calculus or a graphing calculator (it's just a parabola). \nSince the graph of your equation is an inverted parabola, the maximum height of the bullet will occur at the parabola's vertex. Complete the square of h(t) to get h = -18(t - 48)^2  + 41490 or h - 41490 = -18(t - 48)^2. From pre-calculus you should recognize that this is the equation of a parabola of the form y = -18t^2 whose vertex was shifted to (48, 41490). Thus, the maximum height is 41490ft.
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
1146457
vertex (h,k)\nfocus (h,k + p)\n\n(x - h)^2 = 4p(y - k)\n\nvertex : (2,-6)\nfocus : (2,-6 + p)\n\nk + p = -8\n-6 + p = -8\np = -2\n\n(x - h)^2 = 4p(y - k)\n(x - 2)^2 = 4 * -2(y - (-6))\nx^2 - 2x - 2x + 4 = 4 * -2(y + 6)\nx^2 - 4x + 4 = -8(y + 6)\nx^2 - 4x + 4 = -8y - 48\nx^2 - 4x + 8y + 52 = 0\n\nANS : d.) x^2 - 4x + 8y + 52 = 0
1103883
Y=X^5-X^-3+X^2\nY'=5X^4+3X^-4+2X\nY''=20X^3-12X^-5+2
934749
the fomula is:\n\nCxHy + (x+y/4)O2 ---> x CO2 + (y/2)H2O
187414
2^y = 64/y\n\n\nthis is the intersection of two curves:\nt=2^y\nand\nt=64/y\n\nthat is an exponential and an equilateral hyperbola.\nPlotting the graphs you can have an idea of the intersection (in this case you have the exact solution by plotting integer values, but in general you may need to solve by iterative numeric formulas, after finding graphically a rough estimate of the solution)
324456
give parameters for x, y, and z.  Otherwise you cannot prove it, as no parameters are given for the equation as you described it.