Find the equation of the lines having slope -1 that are tangent to the curve y=1x−1,x≠1
The equation of the given curve is y=1x−1,x≠1
The slope of the tangent to the given curve at any point (x,y) is given by
dydx=−1(x−1)2
For tangents having slope =-1, we must have
−1=−1(x−1)2⇒(x−1)2=1⇒x−1=±1⇒x=1±1=2,0
When x=2, then from Eq. (i), we get y=12−1=1
When x=0, then from Eq. (i), we get y=10−1=−1
Thus, the points on the given curve at which slope of tangent is -1, are (2,1) and (0,-1).
∴ Equation of tangent at (2,1) is y-1=-1(x-2)or x+y-3=0
and equation of tangent at (0,-1) is y-(-1)=-(x-0) or x + y + 1 = 0
Hence, the equation of the required line are x+y-3=0 and x + y + 1 = 0