what is a use of an ellipse in real life?
962668
The Capitol in Washington D.C. has a ceiling that follows an elliptical curve.  It is over the original senate chamber.\nThe nature of an ellipse is that the sum of the distances from the two focal points remains constant.  Also, the angles between rays from the foci to the curve will be equal.\nThe guys who built the senate chamber used those principles to create 2 focal points on the senate floor.  If you are sitting at one focal point, you can hear another person at the other focal point as clearly as if he/she is whispering in your ear.\nbtw. race tracks are ovals, not ellipses.
545673
there are lots of medical devices taking ideas from trigonometry.\none simple example is hip prostheses..the angle at which it should be designed implanted and all are decided by trignometric techniques. I can give you more examples if you r really interested
525762
(m)\n\nBY  FOUR-CENTER  METHOD The   four-center   method   is   used   for   small ellipses.  Given  major  axis,  AB,  and  minor  axis, CD, mutually perpendicular at their midpoint, O, as shown in figure 4-45, draw AD, connecting the end points of the two axes. With the dividers set to  DO,  measure  DO  along  AO  and  reset  the dividers on the remaining distance to O. With the difference  of  semiaxes  thus  set  on  the  dividers, mark  off  DE  equal  to  AO  minus  DO.  Draw perpendicular  bisector  AE,  and  extend  it  to intersect the major axis at K and the minor axis extended  at  H.  With  the  dividers,  mark  off  OM equal  to  OK,  and  OL  equal  to  OH.  With  H  as a  center  and  radius  R1 equal  to  HD,  draw  the bottom  arc.  With  L  as  a  center  and  the  same radius as R1, draw the top arc. With M as a center and the radius R2 equal to MB draw the end arc. With K as a center and the same radius, R2, draw the  end  arc.  The  four  circular  arcs  thus  drawn meet,  in  common  points  of  tangency,  P,  at  the ends  of  their  radii  in  their  lines  of  centers
1079158
Focal radii = sum of the distances from a point on the ellipse to F1 and F2\n\nFoci radii is 20, which means that 2a (twice the major axis) is 20.\na = 10\n\nYou can find the centre using the equation for midpoint [(x1+x2)/2, (y1+y2)/2] and the centre will be at (-2, 2.5)\n\nThis means that c (distance between centre and focii) is 4.5\n\nYou can use the equation a^2 = b^2 + c^2 to find b:\nb^2 = a^2 - c^2\nb^2 = 10^2 - 4.5^2\nb^2 = 79.75\n\nSince this ellipse's major axis is on the y-intercept (focii have the same x-coordinates), the equation for it will follow this:\nx^2/b^2 + y^2/a^2\n\n(x+2)2/79.75 + (y-2.5)/100\n\n(I forgot the centres, doy, how stupid!)
806229
First, the equation of the ellipse must be:\nb^2 x^2 + a^2 y^2 = a^2 b^2; You seem to have made a typing mistake.\n\nA tangent to this at (x1, y1) is\n\nb^2 xx1 + a^2 yy1 = a^2b^2\n\nSince it is in the first quadrant, both x1 and y1 are positive.\n\nThe x and y intercepts of this line are\nx-intercept = a^2/x1\ny-intercept = b^2/y1\nThese are also the base and height of the triangle formed by the axes and the line;  the area is\na^2b^2/(x1y1)\nThis will be minimum when x1y1 is maximum.\n>>x1^2y1^2 is maximum\n>> x1^2 (a^2b^2 - b^2x1^2) / b^2 is maximum;(using the equation of the ellipse)\n>>a^2b^2x1^2 - b^2x1^4 is maximum\n>> 2a^2b^2x1 - 4b^2x1^3 = 0\n>> a^2 - 2x1^2 = 0; (we know x1 > 0)\n>> x1 = a/sqrt(2)\n>> y1 = sqrt((a^2b^2 - b^2a^2/2)/a^2)\n= b/sqrt(2)\n\nFinally, the equation of the required tangent is\n(b^2 a/sqrt(2))x + (a^2 b/sqrt(2))y - a^2b^2 = 0
136253
You're already on a website.  What is it with people these days?  Look in your book.  \nInteresting though: the derivative of the area of a circle is equal to the circumference of it.  Works the same for a sphere too.
470236
Ellipse is for positive curvature and hyperbola for negative curvature.
403281
take a straw (cylinder) and ram it through a potato.  If you do it right, the straw will go right through the potato!  The cylinder is the strongest and most stable of the shapes and you can see why soups, veggies, fruits, etc. all come in cans!\n\nRight triangels, or right angles are all around you.  The floor and the wall of you room for example.  The bottom of a pool is a good example of an obtuse triangle, suspension bridges - trapezoid, if you look at the tip of a pencil - it's a cone but the shape of a cone when you look at it head on is a scalene triangle
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| .
234828
A tool to measure diameter of a round object.
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First, you won't have a complete understanding of how i is applied in real life until you study complex analysis. Note, the form of a complex number is: a + bi, where a,b are real numbers. i means a is 0 and b is 1.\n\n     i is used in the applications (conformal mapping) to solve a number of physical problems in heat conduction, electrostatic potential, and fluid flow, to name a few.\n\n     In the study of heat conduction, the FLUX across a surface within a solid body at a point on that surface is the quantity of heat flowing in a specified direction normal to the surface per unit time per unit area at the point. Flux is usually measured in calories per second per square centimeter. The stove in your kitchen has this. Remember the FLUX capacitor in the movie, 'Back to the Future'?\n\nIn an electrostatic force field, the field intensity at a point is a vector representing the force exerted on a unit positive charge placed at that point. The elctrostatic potential is a scalar function of the space coordinates such that, at each point, its directional derivative in any direction is the negative of the component of the field intensity in that direction. Gravitational potential is another example in physics.\n\nComplex harmonic function functions play an important role in hydrodynamics and aerodynamics. We let vectors be in complex form. That is, V = p + iq would represent the velocity of a particle of a fluid at any point. Note, fluid could be liquid or air around the surface, as in aerodynamics.\n\nHope this helps.
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
1053962
I am assuming this is an "I hate math so I'm gonna complain about its lack of relevance" question and not a real question, but I'll answer it anyway.  I can't tell you every use of it, I suppose I've seen several, but here's one I see a lot.  If you have an object thrown in a trajectory, and you need to know when it will cross a certain height, or when it will be going a certain speed (I dunno, in case you're lobbing hand grenades or missiles someday when you're bored), the quadratic equation will come up in the solution, and it will give you BOTH of the answers, one for when the object is on the way up and one for when it's on the way down.
837822
I would have asked the same question when I was in high school, but now that I am in college I see a real use for it. Engineers, mathematicians, scientists, masons, and carpenters use geometry all the time. It's a lot more useful that you think.
425472
There is no simple formula for the perimeter of an ellipse. The exact value is given by an ELLIPTIC INTEGRAL OF THE SECOND TYPE -- in the past people used extensive tables to find approximate answers, but today one gets greater accuracy using a calculator to approximate the integral. When the ellipse has semi-axes of length a and b (so that its equation is x^2/a^2 + y^2/b^2 = 1\nA simple approximate formula for perimetre is\n2pi*sqrt[(a^2+b^2)/2]
474415
Because of the ratio of area inside a shape compared to the outline isnt the most efficient (outline 4a cf area a^2) where a is the length of side for the square (rationalises to 4:a)\n\none would find other shapes to be more efficient like hexagons, with circles having the best ration between area and circumference (PI*d cf (PI*d^2)/2) which rationalises to 1:d/2 where d is the diameter (width) of the circle.\n\nthe circle will contain the maximum area within for the minimum circumference thus rendering it cheaper if you are say making a box for a known quantity of goods.\n\nother than that the answer can depend on what the final calculations are to be used for
385100
The volume of an ellipse is a calculus problem. The fact that it's eggshaped isn't enough. There are hundreds of varying ellipses, all egg shaped.