JEE MAIN - Mathematics (2009 - No. 16)
The shortest distance between the line $$y - x = 1$$ and the curve $$x = {y^2}$$ is :
$${{2\sqrt 3 } \over 8}$$
$${{3\sqrt 2 } \over 5}$$
$${{\sqrt 3 } \over 4}$$
$${{3\sqrt 2 } \over 8}$$
Explanation
Let $$\left( {{a^2},a} \right)$$ be the point of shortest distance on $$x = {y^2}$$
Then distance between $$\left( {{a^2},a} \right)$$ and line $$x - y + 1 = 0$$
is given by
$$\,\,\,\,\,\,\,\,D = {{{a^2} - a + 1} \over {\sqrt 2 }} = {1 \over {\sqrt 2 }}\left[ {{{\left( {a - {1 \over 2}} \right)}^2} + {3 \over 4}} \right]$$
It is min when $$a = {1 \over 2}$$ and $$D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}$$
Then distance between $$\left( {{a^2},a} \right)$$ and line $$x - y + 1 = 0$$
is given by
$$\,\,\,\,\,\,\,\,D = {{{a^2} - a + 1} \over {\sqrt 2 }} = {1 \over {\sqrt 2 }}\left[ {{{\left( {a - {1 \over 2}} \right)}^2} + {3 \over 4}} \right]$$
It is min when $$a = {1 \over 2}$$ and $$D{}_{\min } = {3 \over {4\sqrt 2 }} = {{3\sqrt 2 } \over 8}$$
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