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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Sep 2007 07:06:13 IST
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The equation of the tangent at the point P(t), where t is any parameter, to the parabola y2 = 4ax is
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Hiii... Always Rem'ber... for any second degree curve ax2 + by2 + 2gx +2fy + 2hxy + c = 0, the equation to any tangent at point (x1 , y1) is given by replacing x2 by xx1 y2 by yy1 2x by x+x1 2y by y+y1 2xy by xy1 + yx1 here the curve given is y2 = 4ax and x1 ,= at2 and y1= 2at thrfore, equation of curve at any point P(t) is given by yy1= 2a (x+x1) => 2aty= 2a (x + at2) => ty = x + at2 do knock back..if any problem
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3 Sep 2007 19:09:26 IST
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Differentiate the given parabola and substitute P(at^2,2at).U will
get slope of tangent .Now use point-slope form u will get the
equn., of tangent.
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***T.Venkat*** |
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4 Sep 2007 12:36:31 IST
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Hiii... Always Rem'ber... for any second degree curve ax2 + by2 + 2gx +2fy + 2hxy + c = 0, the equation to any tangent at point (x1 , y1) is given by replacing x2 by xx1 y2 by yy1 2x by x+x1 2y by y+y1 2xy by xy1 + yx1 here the curve given is y2 = 4ax and x1 ,= at2 and y1= 2at thrfore, equation of curve at any point P(t) is given by yy1= 2a (x+x1) => 2aty= 2a (x + at2) => ty = x + at2
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