Question

Find the equations of tangents to y = 1/x-1,x≠1 parallel to the line x+y+7=0

Answer 1

we have to find equation of tangent parallel to y = 1/(x - 1), x ≠ 1 parallel to the line x + y + 7 = 0

we see , slope of line {x + y + 7 = 0} = -1
slope of line should be equal slope of tangent to the curve y = 1/(x - 1) , x≠ 1 [ because tangent is parallel to line ]
so, slope of tangent = -1 ......(1)

now, y = 1/(x - 1)
differentiate with respect to x,
dy/dx = -1/(x - 1)²
so, slope of tangent = dy/dx = -1/(x - 1)²
=> - 1 = -1/(x - 1)² [ from equation (1) ]
=> (x - 1)² = 1
=> x² - 2x = 0
=> x(x - 2) = 0
x = 0, 2
put it in y = 1/(x - 1), x ≠ 1
e.g., y = -1 along x = 0
y = 1 along x = 2

hence, there are two points (0,-1) and (2,1) where tangent of y = 1/(x - 1) , x ≠ 1 is parallel to line x + y + 7 = 0

now equation of tangents :
tangent passing through (2,1) and slope is -1
(y - 2) = -1(x - 1)
x + y = 3

tangent passing through (0,-1) and slope is -1
(y +1) = -1(x - 0)
x + y + 1 = 0

Answer 2

Dear Student:

Given:  y = 1/x-1,x≠1

$m=\frac{dy}{dx} =slope of tangent$

$\frac{dy}{dx}=\frac{d}{dx}(\frac{1}{x-1})$

$\frac{dy}{dx} =\frac{-1}{(x-1)^2}$

tangents  parallel to the line x+y+7=0

y=-x-7

slope of tangent=slope of line=-1

See the attachment: