Question:
The shortest distance between the point \(\left(\frac32,0\right)\) and the curve \(y=\sqrt{x}\), \((x>0)\), is -
The shortest distance between the point \(\left(\frac32,0\right)\) and the curve \(y=\sqrt{x}\), \((x>0)\), is -
Show Hint
Minimise the squared distance \(\left(x-\frac32\right)^2+x\).
Updated On: Nov 5, 2026
- \(\frac{\sqrt3}{2}\)
- \(\frac54\)
- \(\frac32\)
- \(\frac{\sqrt5}{2}\)
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The Correct Option is D
Solution and Explanation
For a point \((x,\sqrt x)\): \(D^2=\left(x-\frac32\right)^2+x=x^2-2x+\frac94\).
\(\frac{d(D^2)}{dx}=2x-2=0\Rightarrow x=1\) (minimum, since the second derivative is \(2>0\)).
\(D^2_{\min}=1-2+\frac94=\frac54\), so \(D=\frac{\sqrt5}{2}\).
Answer: (4) \(\frac{\sqrt5}{2}\)
\(\frac{d(D^2)}{dx}=2x-2=0\Rightarrow x=1\) (minimum, since the second derivative is \(2>0\)).
\(D^2_{\min}=1-2+\frac94=\frac54\), so \(D=\frac{\sqrt5}{2}\).
Answer: (4) \(\frac{\sqrt5}{2}\)
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