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Find the points on the curve `x^2+y^2-2x-3=0`at which the tangents are parallel to the x-axis. |
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Answer» Equation of curve ` x^(2)+y^(2)-2x - 3 = 0`…(1) ` rArr 2x + 2y (dy)/(dx) - 2 =0` ` rArr (dy)/(dx) = (1-x)/y` Slope of tangent at point `(x, y) is m = (1-x)/y` but the tangent is parallel to x-axis. ` :. M = 0` ` rArr (1-x)/y = 0 rArr x = 1` put x = 1 in equation (1) ` 1+ y^(2) - 2 - 3 = 0` ` rArr y^(2) = 4 rArr y = pm 2` `:.` Required point = (1, 2) and (1, -2) |
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