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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The shortest distance between the line $y-x=1$ and the curve $x=y^2$ is

$\begin{array}{1 1}(A)\;\frac{2 \sqrt 3}{8} \\(B)\;\frac{3 \sqrt 2}{5} \\(C)\;\frac{\sqrt 3}{4} \\(D)\; \frac {3 \sqrt 2}{8} \end{array}$

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1 Answer

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Let $(a^2,a)$ be thw point of shortest distance on $x=y^2$
The distance between $(a^2,a)$ and line $x-y+1=0$ is given by
$D= \large\frac{a^2-a+1}{\sqrt 2}$
$\qquad= \large\frac{1}{\sqrt 2} \bigg[ (a- \large\frac{1}{2})^2+\frac{3}{4}\bigg]$
If is min when $a= \large\frac{1}{2} $ and $D_{min}=\large\frac{3}{4 \sqrt 2}$
$=> \large\frac{3 \sqrt 2}{8}$
Hence D is the correct answer.
answered Apr 8, 2014 by meena.p
 

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