A) \[\frac{\sqrt{5}}{2}\]
B) \[\frac{\sqrt{3}}{2}\]
C) \[\frac{3}{2}\]
D) \[\frac{5}{4}\]
Correct Answer: A
Solution :
Let P be the point nearest to \[\left( \frac{3}{2},0 \right),\]then normal at |
P will pass through \[\left( \frac{3}{2},0 \right).\] |
Let Co-ordinates of P be \[s\left( \frac{{{t}^{2}}}{4},\frac{t}{2} \right)\] |
Hence, equation of normal is \[y+tx=\frac{t}{2}+\frac{{{t}^{2}}}{4}\] |
The line passes through \[\left( \frac{3}{2},0 \right)\] |
\[\frac{3t}{2}=\frac{t}{2}+\frac{{{t}^{3}}}{4}\]\[\Rightarrow \]t = 2 \[(-2,0\] are rejected) |
Hence, nearest point is (1 ,1) |
Distance \[\sqrt{{{\left( \frac{3}{2}-1 \right)}^{3}}+{{(1-0)}^{2}}}=\frac{\sqrt{5}}{2}.\] |
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