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Application Of Derivatives

Question
CBSEENMA12035156

Find the equations of all lines having slope -1 that are tangents to the curve straight y equals fraction numerator 1 over denominator straight x minus 1 end fraction. straight x not equal to 1.

Solution

The equations of the curve is straight y equals space fraction numerator 1 over denominator straight x minus 1 end fraction space equals left parenthesis straight x minus 1 right parenthesis to the power of negative 1 end exponent
therefore space space space space space space dy over dx space equals space left parenthesis negative 1 right parenthesis thin space left parenthesis straight x minus 1 right parenthesis to the power of negative 2 end exponent space equals space minus fraction numerator 1 over denominator left parenthesis straight x minus 1 right parenthesis squared end fraction
therefore space space space space space space slope space of space tangent space equals space minus fraction numerator 1 over denominator left parenthesis straight x minus 1 right parenthesis squared end fraction
But space slope space of space tangent space space equals space minus 1 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis given right parenthesis
therefore space space space space space space minus fraction numerator 1 over denominator left parenthesis straight x minus 1 right parenthesis squared end fraction space equals space minus 1 space space space space rightwards double arrow space space space space space fraction numerator 1 over denominator left parenthesis straight x minus 1 right parenthesis squared end fraction space equals space 1 space space space space rightwards double arrow space space space space left parenthesis straight x minus 1 right parenthesis squared space equals space 1
rightwards double arrow space space space space space space straight x minus 1 space equals space plus-or-minus 1 space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space straight x space equals space 0 comma space space 2
When space straight x space equals space 0 comma space space space space space straight y space equals space fraction numerator 1 over denominator 0 minus 1 end fraction space equals space minus 1
When space straight x space equals space 2 comma space space space space straight y space equals space fraction numerator 1 over denominator 2 minus 1 end fraction space equals space 1 over 1 space equals space 1

∴   there are two tangents to the given curve with slope – 1 and passing through the points (0, – 1) and (2, 1).
The equation of tangent through (0, – 1) is
y – (– 1) = – 1 (x – 0) or y + 1 = – x  or  x + y + 1 = 0
The equation of tangent through (2, 1) is
y – 1 = – 1 (x – 2) or y – 1 = – x + 2 or x + y – 3 = 0