Application of Derivatives

Question

Find the equation of all lines having slope 2 and being tangent to the curve $straight y plus fraction numerator 2 over denominator straight x minus 3 end fraction space equals space 0$ .

Solution

The equation of curve is
$straight y plus fraction numerator 2 over denominator straight x minus 3 end fraction space equals space 0 space space space space space or space space space space straight y space equals negative fraction numerator 2 over denominator straight x minus 3 end fraction$
or $straight y equals negative 2 left parenthesis straight x minus 3 right parenthesis to the power of negative 1 end exponent$
$therefore space space space space space space space space space space space dy over dx space equals space left parenthesis negative 2 right parenthesis thin space left parenthesis negative 1 right parenthesis thin space left parenthesis straight x minus 3 right parenthesis to the power of negative 2 end exponent space equals space fraction numerator 2 over denominator left parenthesis straight x minus 3 right parenthesis squared end fraction therefore space space space space space space slope space of space tangent space space equals space fraction numerator 2 over denominator left parenthesis straight x minus 3 right parenthesis squared end fraction But space slope space of space tangent space space equals space 2 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 fraction numerator 2 over denominator left parenthesis straight x minus 3 right parenthesis squared end fraction space equals space 2 space space space space space space space space space space space rightwards double arrow space space space space space space space fraction numerator 1 over denominator left parenthesis straight x minus 3 right parenthesis squared end fraction space equals space 1 space space space space space space space rightwards double arrow space space space space left parenthesis straight x minus 3 right parenthesis squared space equals space 1 therefore space space space space space space space space space space space space straight x minus 3 space equals space minus 1 comma space space 1 space space space space space space space space space space rightwards double arrow space space space straight x space equals space 2 comma space 4 When space straight x space equals space 2 comma space space space straight y space equals space minus fraction numerator 2 over denominator 2 minus 3 end fraction space equals space minus fraction numerator 2 over denominator negative 1 end fraction space equals space 2 When space straight x space equals space 4 comma space space space straight y space equals space minus fraction numerator 2 over denominator 4 minus 3 end fraction space equals space minus 2 over 1 space equals space minus 2$
there are two tangents to the given curve with slope 2 and passing through the points (2, 2) and (4, -2).
The equation of line through (2, 2) is
y – 2 = 2 (x – 2) or y – 2 = 2x – 4 or 2 x – 2 = 0
The equation of line through (4, – 2) is
y – (– 2) = 2 (x – 4) or y + 2 = 2 x – 8 or 2 x – y – 10 = 0

For Daily Free Study Material Join wiredfaculty