Question:
The shortest distance between the point $\left(\frac{3}{2}, 0\right)$
and the curve $y=\sqrt{x},(x>0)$ is :
Correct Option: 1
Solution:
Let points $\left(\frac{3}{2}, 0\right),\left(t^{2}, t\right), t>0$
Distance $=\sqrt{\mathrm{t}^{2}+\left(\mathrm{t}^{2}-\frac{3}{2}\right)^{2}}$
$=\sqrt{t^{4}-2 t^{2}+\frac{9}{4}}=\sqrt{\left(t^{2}-1\right)^{2}+\frac{5}{4}}$
So minimum distance is $\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$
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