Hyperbolas are called Confocal if they have the same foci. Show that the hyperbolas...?
920780
The general formula for a North-South lying hyperbola is\n(y-y0)^2 / b^2  - (x-x0)^2 / a^2 = 1.\nSuch an hyperbola has foci (x0,y0+-c) where c^2=a^2+b^2.\n\nIn this case x0=0, y0=0, a^2 = 16-k, b^2=k.\nNote that a^2 and b^2 are defined only for 0<k<16. \nAssuming the latter is true, we have: this leads to c^2 = 16, c=+-4. \nTherefore the foci are *all* located at (0,+-4) independent of k.
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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)
45435
commmon one is Convex lens\n(with different angles, to magnify upto diff levels)\nbut for higher levels, it depends upto the magnifying object and distance also
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No asymptote.  It just keeps growing. That is, as x─►±∞, y ─►±∞.
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Is parabola(rational funtion) a polynomial?\n\nok, you know that a parabola is not a rational function. a parabola is the graph of a quadratic function that looks like:\n\nf(x) = y = ax^2 + bx + c. \n\na polynomial is a sum of finite monomials in which coefficients of terms are real numbers and exponents of variables are whole numbers.\n\nthus, a quadratic is a polynomial.\n============================================\n\n\nnow, is hyperbola a polynomial?\n\nwhat does a hyperbolic function look like?\n\n(x-h)^2/a^2 - (y-k)^2/b^2 = 1\n\nif you solve for y, we do not have a polynomial.
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In optics and physics, a wavefront is a line of points (called a locus) having the same phase. For a wave propagating in 3 dimensions, the locus is a surface.
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Hi,\nThe answer is C.\n\nYou could make a ray-diagram and see for yourself...
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Light is refracted from a single point. Since the light which gets refracted into the same colors is all the same distance from the point source, a circle is formed.\n\n-Wied
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A parabola is a shape formed by all the points that are an equal distance from a line, called the directrix, and a point, called the focus. The interesting thing about parabolas is if you have a three-dimensional figure formed by rotating a parabola on its vertex, if you place a light source at the focus, the light that shines back and hits the parabolic surface will be reflected outward in parallel lines, creating a strong beam. This is how flashlights work. Also, if you use a parabolic shape for an antenna, signals coming in from space or whatever will hit the surface of the parabola and all be reflected onto the focus. If you put a receptor there, you will receive a very strong signal. That's a couple real life uses of parabolas.
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http://mathworld.wolfram.com/PoincareConjecture.html
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http://www.math.niu.edu/~beachy/aaol/frames_index.html
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Yes!!!!!!!!! Email me, and I'll try to shed more light on this. Parabolas are where my genius comes out :)
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The equation is most likely not a parabola. In parabola equations, only one variable is squared. However, if A or B = 0, then you have a sideways or vertical parabola, respectively.\nEx: x^2+5x+6=y\n\nThe equation could potentially be a circle if A=B. This is because circles are not distorted in the x or y direction; they are perfectly round.\nEx: x^2+2x+1+y^2+4y+4=9 or (x+1)^2+(y+2)^2=3^2.\n\nYour equation is most characteristic of an ellipse, but only assuming so if A and B are different, positive numbers. Ellipses are bent in either the x or y direction.\nEx: 9y^2+36y+36+25x^2+50x+25=225\n\nLast but not least, your equation could describe a hyperbola, where either A or B (but not both) is negative.\nEx: 9y^2+36y+36-(25x^2+50x+25)=225
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I have no idea why they are so called, but a tesselating pattern is one which repeats with no gaps. Tiled floors are a good example, from the basic squares to the more complicated tiled floors seen elsewhere
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I don't believe it can.  This is an example of a unicursal problem and you can find some more details at the puzzle link in the reference below, you'll need to scroll down to puzzle number 51.\n\nFrom my days of O Level Maths I remember that we touched on these and I seem to recall that there was an equation to do with the number of vertices and nodes that a shape had which would determine if it was unicursal or not.\n\nBasically only unicursal shapes (those with a specific ratio of nodes and vertices) can be drawn without removing a pencil from the page and without tracing over an existing line.\n\nI'll leave further research to you.\n\nHave fun
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No. xy = k is an hyperbola in 2D. xy = kz is an hyperboloid on 3D.
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"Well, here's a little bit about parabolas, but not the definitive treatise on the subject:\n\nThere are two equivalent ways of looking at them (a little algebra will turn one method into the other):\n\n1. The graph of a second-degree polynomial function in one variable(known as the old y = ax^2 + bx + c rule usually seen in first or second year algebra class).\n\n2. The set of all points that at the same time are equidistant from a given point and a given line (the point is the focus, and the line is the directrix).\n\nNow, if you know the vertex of the parabola, it's got to be exactly midway between the focal point and the directrix line (equidistant).\n\nOne of the interesting things about a focal point is that the distance across the parabola through the focus is equal to four times the distance from the focus to the vertex.  This isn't too tough to show, those side points on the parabola must be twice that distance each to \nthe directrix, and therefore, to the focus. The arc isn't really circular.  It's more like an infinite valley or mountain (depending on whether its vertex, or "turning point" is at the top or bottom).  The steepness of the curve gets more steep the further away from the \nvertex, near which the curve *looks* more round before starting to straighten out....not that it actually straightens out, but it is less obviously curved further away from the vertex."