On the ellipse 4x^2+9y^2=1,the points at which the tangents are parallel to the line 8x=9y are?
488924
since the tangent is parallel to line 8x=9y... the relation 8x=9y holds good for the tangent so\nx=9/8 y\nsince the tangent touches the ellipse the points must be on the ellipse so substitute x=9/8 y in the equation of the ellipse 4x^2 + 9y^2 = 1\n\n4(9/8 y)^2+ 9y^2 = 1\ny^2(81/16 + 9) =1\nY^2(225/16)=1\ny^2=16/225\ny=+(or)- 4/15\ntherefore \nx= 9/8 (+(or)- 4/15)\nx=+(or)- 3/10\ntherefore the points are (4/15,3/10) and another tangent parallel to the previous on at (-4/15,-3/10)
197260
the graph of g(x)= y = x-2x^2 is a curve, on differntiating this\nequation we get 1-4x=slope of line\nd/dx g(x)\nbut slop of line is tan(0) since it is parallel to x axis, so 1-4x=0, so x=.25, now substitute x=.25 in g(x), we get y=0\nso, the point is (.25,0) or (1/4,0)
597648
y= -4x^2+8x-1\ny = -(4x^2-8x+1)\ny = -[(2x-2)^2 - 4 + 1]\ny = -[(2x-2)^2 - 3]\ny = 3 - (2x-2)^2\n\nSo the vertex is (1,3) and it is a maximum point.
846578
1.  Since f(x) = x² + bx + c, and you know f(4) = 9 (since (4,9) is a point on the parabola, specifically the vertex), you can solve for f(4) = \n\n16 + 4b + c = 9, so 4b + c = -7\n\nSince (4,9) is the vertex, you know the tangent at this point is zero, i.e., f '(4) = 0.  (I hope this is a calculus HW problem so that you know what I am talking about!)\n\nSo get f '(x) = 2x + b\n\nSince f '(4) = 0, we have 2(4) + b = 0, so b = -8\n\nSince b = -8, and 4b + c = -7, we can now solve for c:\n\n4(-8) + c = -7, so c = 25\n\n----------\n\n2.  Not sure I can help here.  I took the first and second derivatives of h(x) to determine that the function has a local min at x=7/3 and a local max at x=-1, but h(7/3) ≠ -5 and h(-1) ≠ 2, so not sure where to go from here, sorry!\n\n----------\n\n3.  A parabola has a 2nd-degree polynomial equation, or y = ax² + bx + c.  We know two values, namely (4,4) and (2,8) so we can set up two equations like thus:\n\n4 = a(4²) + b(4) + c\n8 = a(2²) + b(2) + c\n\nwhich we can rewrite as:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n\nRight now we have 3 unknowns and only 2 equations, but we can get a third equation by taking the first derivative, or f '(x) = 2ax + b.  Since the vertex is at (4,4) we know that f '(4) = 0, and we can get a third equation:\n\n0 = 2a(4) + b = 8a + b\n\nSo now we have 3 equations in 3 unknowns and we can solve for a, b, and c:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n0 = 8a + b\n\nSubtract equation 1 from equation 2, then divide by 2 to get:  \n2 = -6a - b\n\nNow add equation 3 to this to get 2 = 2a, or a=1.\n\nNow let's back-substitute to get b and c:\n\n2 = -6(1) - b, so b = -8\n\n8 = 4a + 2b +c\n8 = 4(1) + 2(-8) + c, so c = 20\n\nSo the equation of our parabola is:\nf(x) = x² -8x + 20\n\n(Double check to make sure f(4)=4 and f(2)=8).
567898
16x²-9(y+16)²-144=0\n=>16x²-9(y+16)²=144\n=>x²/9-(y+16)²/16=0\n=>(x/3-y/4-4)(x/3+y/4+4)=0\nsolving for 2 eqn.\nx/3-y/4-4=0 ....[1]\nx/3+y/4+4=0 ....[2]\nAdding both eqn\n2x/3=0 so x=0\nPutting x in eqn 1\ny/4=-4\ny=-16\nSo values of x and y are x=0,y=-16
895342
x^2-9x-x+9=0\n(x-9)(x-1)=0\nx=9 or x=1
104006
Start with \n\n    4x + 9y = 24\n\nsubstitute the value of y from the second equation:\n\n    4x + 9(-x/3 + 2) = 24\n--> 4x -3x + 18 = 24\n--> x = 24 - 18\n--> x = 6\n\nand now y can be computed,\n\n    y = -6/3 + 2\n--> y = 0
968996
y^2-2xy-8=0...........(1)\nor, 2y(dy/dx) - 2 (x dy/dx + y) = 0\nor, y(dy/dx) - x(dy/dx) = y\nor, dy/dx = y/(y-x)\n\nLet us choose an arbitrary pt (1,k) on st line x=1\n\nhence, dy/dx(at (1,k) = k/(k-1).\nequation of the tangent on (1) at (1,k) is,\n\n(y-k) = k/(k-1) (x-1)\nor, (y-k)(k-1) = k (x-1)\nor, yk -k^2 -y +k = kx -k\nor, kx + y(1-k) = 2k - k^2................(2)\n\nfrom (1),\nputting x =1\ny^2-2y-8=0\nor(y-4)(y+2) = 0\nor, y = 4,-2\nhence k can assume the values 4 & -2\nputting the values of k in (2)\n4x + y(1-4) = 2*4 - 4^2\nie, 4x - 3y + 8 =0\n\n&\n-2x + y(1-(-2)) = 2*(-2) - (-2)^2\nie, 2x - 3y = 0\n\nHence the equations are 4x - 3y + 8 =0 & 2x - 3y = 0
912852
Let P be (a,b). The slope of the tangent at P is found from the derivative of the curve: -128 / x³, giving the equation of the tangent as\n\n          y - b = ( -128 / x³ ) ( x - a )\n\nNow find where this tangent lines cuts the X and Y axes. Let x=0 and solve for y. Let y=0 and solve for x to get\n\n          ( 0 , b + 128 / a² ) and ( a + a³b / 128 , 0 )\n\n(a, b) is on the curve so b = 64 / a². Use this to simplify the two intercepts to\n\n          ( 0 , 192 / a² ) and ( 3/2 a , 0 )\n\nUse Pythagoras to find the length between these points as the square root of\n\n          f(a) = 192² / a^4 + 9 / 4 a²\n\nNow the problem is to minimize this expression (ignoring the square root, because it is an increasing function). This is a standard minimization calculus problem. Set f '(a) = 0, and check f "(a) > 0 to find that the minimum occurs when a = 4√2, so the required point is \n\n          ( 4√2 , 2 ).
645941
4x + y >= 4\ny >= -4x +4\n\n-------\ny = 0\n0 >= -4x + 4\nx = 1\n-------\nx = 0 \ny >= 0 + 4\ny = 4\n-------\n\n\nTherefore, shade the region above the line.\n\nTwo points on the line are:\nx-intercept: (0, 4)\ny-intercept: (1, 0)
867445
9x²+16y²=144\n\nDifferentiate with respect to t\n18x * dx/dt + 32y * dy/dt = 0\nSo 18x * 4y + 32y * dy/dt = 0\nie dy/dt = - 72xy/32y\n= -9x/4\n\nd²y/dt² = -9/4*dx/dt = -9y = 0 for stationary points\nie y = 0 \n\nWhen y = 0, x = ±4\n\nSo maximum dy/dt = -9*(-4)/4 = 9\n(and minimum when dy/dt = -9*(4)/4 = -9)
1078325
Let as suppose that your ellipse is rotated by 45 degrees to right (or by -45 degrees). The new coordinates are X = x cos (-45) - y sin (-45) and Y = x sin (-45) + y cos (-45) and they are satisfying X^2/a^2 + Y^2/b^2=1. There for (x + y)^2/2a^2 + (x - y)^2/2b^2 = 1 (in a few easy steps). The mayor axis over the x=y is given by |a| > |b| .
305313
4x^3y^2-16y^3-9y=\ny(4x^3y-16y^2-9)
887097
y=1 is a straight line parallel to x axis\nx=(-1/4) is a straight line parallel to y axis
762369
Find the coordinates of the points where the straight line meets 'y=1' meets the circle 'x^2+y^2-14x+2y+45=0'.\n\n'x^2+1^2-14x+2(1)+45=0\nx^2 -14x +48 = 0\nthe solutions are:\nx=(14 +- sqrt( 14^2 -4(48) ) /2\nx=14+- sqrt(4) )/2\nx= (14 +2)/2 or x= (14-2)/2\nx=8 or x = 6\n.
697523
For any line through 2 points (x-x1)/(y-y1)=(x2-x1)/(y2-y1); thus (x-0)/(y+5)=(8-0)/(9+5) or\nx=(y+5)*4/7 or 7x-4y-20=0 or 4y-7x+20=0 or y=1.75x-5 all are the same!
897443
2x + y = - 9\n\n2x + y - y = - 9 - y\n\n2x = - 9 - y\n\n2x/2 = - 9/2 - y/2\n\nx = - 9/2 - y/2\n\n- - - - - - - -\n\ny values\n\n2x + y = - 9\n\n2x + y - 2x = - 9 - 2x\n\ny = - 9 - 2x\n\n- - - - - - - - - - - - -\n\n2).\n\nx values\n\n-x - y = - 8\n\n- x - y + y = - 8 + y\n\n- x = - 8 - y\n\n- 1(- x) = - 1(- 8) + (- 1)(y)\n\nx = 8 - y\n\n- - - - - - - - - - - -\n\ny values\n\n- x - y = - 8\n\n- x - y + x = - 8 + x\n\n- y = - 8 + y\n\n- 1(- y) = -1(- 8) + (- 1)(x)\n\ny = 8 - x\n\n- - - - - - - - - - - -\n\n3).\n\nx values\n\n- 3x + 4y = 2\n\n- 3x + 4y - 4y = 2 - 4y\n\n- 3x = 2 - 4y\n\n- 3x/ - 3 = 2/- 3 - 4/- 3\n\nx = - 2/3 + 4/3\n\n- - - - - - - -\n\ny values\n\n- 3x + 4y = 2\n\n- 3x + 4y + 3x = 2 + 3x\n\n4y = 2 + 3x\n\n4y/4 = 2/4 + 3x/4\n\ny = 1/2 + 3x/4\n\n- - - - - - - - - -\n\n4).\n\nx values\n\n- 2x - y = - 3\n\n- 2x - y + y = - 3 - y\n\n- 2x = - 3 - y\n\n- 2x/ -2 = - 3/- 2 - y/-2\n\nx = 3/2 + y/2\n\n- - - - - - - - - \n\ny values\n\n- 2x - y = - 3\n\n- 2x - y + 2x = - 3 + 2x\n\n- y = - 3 + 2x\n\n- 1(- y) = -1 (- 3) + (- 1) (2x)\n\ny = 3 - 2x\n\n- - - - - - - -s-