A) greater than 4
B) less than 2
C) greater than 3
D) greater than 2
Correct Answer: B
Solution :
Shortest distance between |
\[{{y}^{2}}=x-2\]and \[y=x\] |
\[\frac{dy}{dx}\]at point P will be 1 |
Differentiating the curve |
\[2yy'=1\]\[\Rightarrow \]\[y'=\frac{1}{2y}=\frac{1}{2\alpha }=1\] |
\[\therefore \] \[P\left( \frac{9}{4},\frac{1}{2} \right)\] |
minimum distance\[=PQ\] |
\[=\left| \frac{\frac{9}{4}-\frac{1}{2}}{\sqrt{2}} \right|=\frac{7}{4\sqrt{2}}.\] |
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