Find the vertices, foci, and eccentricity of the ellipse? Its different compared to others. Please help?
959248
When you divide by 3, you get 2/3 x^2 + 1/3 y^2 = 1\n\nAs fractions with just x^2 and y^2 on top, this becomes\n\nx^2 / (3/2) + y^2 / 3 = 1\n\nSo a = 3 and b = 3/2.  Use these to find c.\n\nThen use the formulas to get the rest.  Just checking my memory, is eccentricity c/a?
794182
What nicknamr said, and also, the vertex would be located at the x-value found for the axis of symmetry, and the y-value for that x-value.  So, -b/(2a) = -12/(2 x -1) = 6.  Then, y= -(6)^2 + 12(6) -32 = -36 + 72 -32 = 4.  The vertex comes out to be (6,4).  (I would have answered all the other stuff too, but he already did it, and I found it rather redundant to repeat it all, so I just added on to it =)  )
970313
r=radius\nd=diameter\n\n1 C=2r(pi)\n2 C=d(pi)\n\n(Same thing because 2r=d)
415440
r decreases becuase the ellipse is becoming thinner i think that would happen\n\nhope u find out if my answer is wrong GOOD LUCK!!!!!!!!
700031
You are looking for an ellipse.  The inside of the ellipse would be the area that the scout could survey.  20 would be the distance between the foci and 40 would be the vertices.  Good Luck
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
745252
1) All you need is the circumference, which is pi * diameter, to get the linear distance traveled per revolution.  Then just divide by 100,000 to get it in terms of km, and multiply it by 60 to get it in terms of "per hour".\n\n2) Are you sure you copied this one correctly? The way I figure it, AD = ½ of AB, BE is ½ of BC and CF is ½ of CA.  This means that the left side can be written "4(½AB + ½BC + ½CA).  You then factor out the ½, and you're left with 2(AB + BC + CA), which will always be less than 3(AB + BC + CA)...\n\n3) Start by graphing a right triangle on an X-Y graph with one of the non-right angles at the origin (0,0). Then figure the circle that is formed by the three points of the triangle. When you are finished, the hypotenuse should be the same as a diameter of the circle.\n\nI hope this helps.
401455
Sorry.  This is going to be a little rough, typing the answer.  Here it goes:  \n(x-2)2(squared)+ (y-4)2 (squared)= 25\n\nYou find the midpoint between the two points for the center.  You find the distance between the two points for the diameter and divide that in half for the radius which equals 5.  The center is (2,4).  You square the radius and that is what comes after the equals sign.\n\nThe guy who entered before me is wrong.  I promise.
796855
if you mean an ellipse....\n\narea is equal to pi * minor radius * major radius
919538
it is narrow because a is more than 1.\nThe vertex=1,-1 you take the opposite of h which=x and y=-1\ndo a t table to find the max and min \nput the vertex in and put points for x and solve for y for each one using the equation
910387
The nearest vertex will be any of the four lower vertices because the ball is closer to the bottom than to the top of the cube, and the centre of a square is equidistant from all four vertices.\n\nPicture the right triangle whose vertices are:\nA = the center of the ball\nB = one of the lower vertices\nC = the center point of the bottom face of the cube.\n\nYou want to know the length of segment AB which is the hypotenuse of this triangle.  You know that AC is 6 cm long.  And BC is half the diagonal of the bottom face so it measures 1/2 * sqrt(2) * 14 = 7*sqrt(2) cm.\n\nSo AB = sqrt(AC^2 + BC^2)\n= sqrt(6^2 + (7*sqrt(2))^2)\n= sqrt(36 + 49*2)\n= sqrt(134) cm.
900114
hi bob!\nDo you mean ellipse? If so then its equation is: (x/a)^2+(y/b)^2=1, a=4, b=3;\nHammer nails in ellipse focuses with coordinates [-f 0] & [+f 0], now attach the rope ends to these nails, let the rope length be L, keep the rope taught with pencil and draw.\nLet y=0, then x=a (or x=-a which does not matter here), then L=(a+f)+(a-f)=2*a\nLet x=0, then y=b, then f^2+b^2=(L/2)^2, f=sqrt(a*2-b^2)
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
896595
http://local.wasp.uwa.edu.au/~pbourke/geometry/2circle/\n\nit tells you how
24869
List is a bit long, so heres the link:\nhttp://en.wikipedia.org/wiki/List_of_geometric_shapes
762307
This (rectangular) hyperbola is in the asymptotic equation form.\nHere a=b.\nAsymptotes are the axes! x=0 and y=0\n\nThe general equation is xy = (a^2 + b^2)/4 = c^2/4\n\nc^2/4 = 100\nc^2 = 400\nc=20\nDistance from origin to foci is 20 along the line y=x.\nSo if the foci are at (p,q), and since y=x, p=q then\np^2 + q^2 = 20^2\np^2 + p^2 = 400\n2p^2 = 400\np^2 = 200\np = ±√200 = ±10√2\nq = p =  ±10√2\n\nfoci at (10√2, 10√2) and (-10√2, -10√2)
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?