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Find the equation of tangents to the curve `y=cos(x+y),-2pilt=xlt=2pi`that are parallel to the line `x + 2y = 0`. |
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Answer» Equation of the line parallel to the tangent is, `x+2y = 0 => y = -1/2x` `:.` Slope`(m) = -1/2` So, slope of the tangent will be `-1/2`. Now, `y = cos(x+y)` `=>dy/dx = -sin(x+y)(1+dy/dx)` As `m = -1/2, :. dy/dx = -1/2` `=> -1/2 = -sin(x+y)(1-1/2)` `=> sin(x+y) = 1` `=>x+y = pi/2 or x+y = (-3pi)/2` As, `y = cos(x+y)` `:. y = cos(pi/2) = 0 or y = cos((-3pi)/2) = 0` `:. x = pi/2 or x = (-3pi)/2` So, Equation of the line, when `x = pi/2` `(y-0) = -1/2(x-pi/2)` `=> y = -1/2x+pi/4` `=>4y+2x-pi = 0` Equation of the line. when `x =(-3 pi)/2` `(y-0) = -1/2(x+(3pi)/2)` `=> y = -1/2x-(3pi)/4` `=>4y+2x+3pi = 0` |
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