Equation of an ellipse?
822973
Since the foci are along the y-axis, the major axis of the ellipse will be along the y-axis.  Also, the center of the two foci is the center of the ellipse, which will be (0,0) or the origin.\n\nSo the only thing left to figure out is the length of the minor axis.\n\nThe major axis, minor axis, and focal length (distance from the foci to the center) are related by the Pythagorean theorem:\n\na² + b² = c²\n\nYou know that a = 26/2 = 13, c = 12, so b = 5.\n\nTherefore the equation is:\n\nx²/25 + y²/169 = 1
641717
1 mIn going to get calculator....\nOk i cant find my calc, but here is the answer. \nπ*2+π*3+2π*12+2*√(5^2+18^2) + 2*√(4.5^2+20^2)\nHeres each part\nThe belts touching the pullys\nπ*2+π*3+2π*12\nThe belt in the streight runs\n2*√(5^2+18^2) + 2*√(4.5^2+20^2)\nThe √( is supposed to be square root
930129
r = √1² + (-1)² = √2\nθ = arctan(1/-1) = -π/4 ± n2π, where n is any integer.
585612
I don't believe that's the correct formula approximation.\n\nit should be\n\np = pi Sqrt(2(a^2 + b^2) - 0.5(a-b)^2)\n\nwhere a is major(large radius)\nand b = small radius\n\nin your case a = 4/2 = 2 and b = 3/2 =1.5\n\nSo for your case p = 11.16 inches\n\nwhich makes sense since a circle with a diameter of 4 has a circumference of 12.57 and one with diameter of 3 has a circumference of 9.42. So for your ellipse p would have to be between these two values
274298
Use implicit differentiation to find dy/dx. Standard differentiation rules apply, except when you differentiate a term with a "y" in it you must tack on a dy/dx, so:\n\n2x+8y(dy/dx)=0\ndy/dx = -x/4y\n\nThe tangent line intersects the ellipse at (x1,y1), so the slope of the tangent line will be -x1/(4*y1).\n\nWe can also compute the slope of the tangent line by plugging the two points (-5,0) and (x1,y1) into the standard slope formula m = delta(y)/delta(x) and get that the slope of the tangent line is y1/(x1+5). Since these expressions both describe the slope of the tangent line, they must be equal, so\n-x1/(4*y1)=y1/(x1+5)\nx1^2+5*x1=-4*y1^2\nx1^2+4*y1^2+5*x1=0\n\nSince (x1,y1) is a point on the ellipse, it must satisfy the equation of the ellipse, so x1^2+4*y1^2=5. Substituting this in to x1^2+4*y1^2+5*x1=0 gives us 5+5*x1=0, and x1=-1. At this point, there are actually two y1 values that will satisfy the equation of the ellipse, y1 could be either plus or minus 1 (get these by letting x=-1 in the original equation of the ellipse), and so there are actually two points on the ellipse where the tangent line passes through (-5,0), and those points are (-1,1) and (-1,-1).
914363
Find the center-radious form of the equation of the circle with center = (2,-6) and the tangent to the y-axis.\n\nIf tangent to the y - axis then the point (0, -6) is the point of contact (as tangent to circle _|_ radius at point of contact)\n\nTherefore radius = 2\n\nTherefore required circle is\n\n(x - 2)² + (y + 6)² = 2²
538371
http://en.wikipedia.org/wiki/Cubic_equation
750847
you do not even need to complete the square,\nyour equation is the one of an ellipse\n\n9x^2-36x+4y^2=0\n9(x^2-4x+4)+4y^2=0\n9(x-2)^2+4y^2=36\ndividing by 36\n(x-2)^2/4 + (y^2)/9 =  1\n:D
388285
equate the two equations: \nsin 2x = pi/4 \n2x = inverse sine pi/4 + k2pi or (pi - inverse sine pi/4) + k2pi\n2x = 0.90 + k2pi or 2.24 + k2pi\nx = 0.45 + kpi or 1.12 + kpi\nx = 0.45, 1.12, 3.59, 4.26 ( k = 0,1)
143223
a) (6x-7y)(6x+7y)\nb) y(4xy-7)\nc) (x-8)(x+3)
1093431
what is d2 here..if its line what is its equation?
510711
Look at the triangle formed by the line segment AB.  The diagonal side(I'm assuming it's AB) is the hypotenuse.  Label the side across from θ as D.  Label the straight side next to θ as E.  \ncscθ equals the hypotenuse divided by D\ncscθ=(AB)/D\nI don't know what m is supposed to be in the equation, so I can't help you there.\n\nEdit: To find m, take the inverse tangent of D/E.
109151
It looks like your equation got cut off, but I think you were writing\n\nR = (V^2/2g)(sin (2 theta) +sin (2 theta))\n\nwhich can be immediately simplified, by adding the sin (2 theta) terms, as \n\nR= (V^2/2g )(2)(sin(2 theta) = (V^2/g )sin (2 theta)\n\nwhich is the formula that the other correspondent already gave.  To solve it, simply multiply both sides by g, divide by V^2, and you get \n\n(R g)/v^2 = sin (2 theta)\n\n[Arcsin (Rg/V^2) ]/2 = theta
924424
The equations:\n(a+b)^2=a^2+2ab+b^2\nA^2-B^2=(A+B)(A-B)\nJames Chan's solutions were right.
207955
u need more info than just the length, but here is something which might help u are speaking of ovals( called ellipses in math)\n\n\nwell the formula for them is\n\n\n(x^2)/a^2  +  (y^2)/b^2  =1\n\nwhere a and b are half of the major and minor axis , you only know 1 of these meassures, thats not enough sorry chap, wish i could help, but i think no one can, with only that info
247800
it is basically:\na2 + b2 = c2   -->  (A squared)+(B squared)=(C squared)\n  \n\nif you have a right triangle.you have three sides.the two legs.and the hypotenus(sp?).a and b are the two legs and c is the hypotenus.\n\nin an equation.if you are given a and b.you can square each of them and then add the two values together.the will give you c squared.to find c you then need to take the square root of the value(a squared + b squared).\n\nif you have c and a leg(example:a).you square c.and then subtract a squared from that value(c squared).that will give you b squared.to get b you just get the square root of c squared - a squared.\n\n\nhope that helps.
1033875
What "ChinNa90" gave was a very general form of a conic section equation - it does not have to be only circle, but also a parobola or ellipse or hyperbola. \n\nThe standard and easily understandable form of a circle in 2D Cartesian (x and y) coordinates is given by:\n\n(x - a)^2 + (y - b)^2 = r^2\n\nwhere:\nPoint (a,b) is the Centre of the circle\nr is the Radius\n\nTherefore, if you wanted the equation of a circle whose centre is (3,4) and radius is 7 units, then it's equation would be:\n\n(x - 3)^2 + (y - 4)^2 = 7^2\n\nx^2 + y^2 - 6x - 8y + 25 = 49\n\nx^2 + y^2 - 6x - 8y - 24 = 0