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Find the equation of all line having slope - 1 thatare tangent to curve y=1/x-1. xnot equal to 1 |
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Answer» It is given that y = 1/(x - 1) , x ≠ 1 differentiate y with RESPECT to x, dy/dx = -1/(x - 1)² , x ≠ 1 so, slope of TANGENT of the curve at (x,y) = dy/dx = -1/(x - 1)² A/C to QUESTION, we equation of all lines having slope -1 that are TANGENTS to the curve, y = 1/(x -1) , x ≠ 1. so, slope of tangent = - 1 => -1/(x - 1)² = -1 => (x - 1)² = 1 square root both sides, x - 1 = ± 1, x = 1 ± 1 = 0, 2 put x = 0, in the curve y = 1/(x - 1) y = 1/(0 - 1) = -1 now equation of line passing through (0,-1) : (y + 1) = -1(x - 0) => y + 1 + x = 0 hence, equation is x + y + 1 = 0 x+y+1=0 put x = 2, in the curve y = 1/(x - 1) y = 1/(2 - 1) = 1 now equation of line passing through (2,1) : (y - 1) = -1(x - 2) => y - 1 + x - 2 = 0 => x + y - 3 = 0 hence, equation is x + y - 3 = 0 x+y−3=0 |
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