Engineering
Mathematics
Maxima and Minima
Question

The shortest distance between line y – x = 1 and curve x = y2 is :

\(\frac{{\sqrt 3 }}{4}\)

\(\frac{{3\sqrt 2 }}{8}\)

\(\frac{8}{{3\sqrt 2 }}\)

\(\frac{4}{{\sqrt 3 }}\)

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Solution

y – x = 1

y2 = x

                            \(2y\frac{{dy}}{{dx}} = 1\)

                           \(\frac{{dy}}{{dx}} = \frac{1}{{2y}} = 1\)   

                                  \(y = \frac{1}{2}\)

                              \(x = \frac{1}{4}\)

tangent at                              \(\left( {\frac{1}{4},\frac{1}{2}} \right)\)

                               \(\frac{1}{2}y = \frac{1}{2}\left( {x + \frac{1}{4}} \right)\)

                             \(y = x + \frac{1}{4}\)

                                   \(y - x = \frac{1}{4}\)

distance                              \( = \left| {\frac{{1 - \frac{1}{4}}}{{\sqrt 2 }}} \right| = \frac{3}{{4\sqrt 2 }} = \frac{{3\sqrt 2 }}{8}\)