JEE PYQ: Application Of Derivatives Question 71
Question 71 - 2019 (10 Jan Shift 1)
The shortest distance between the point $\left(\frac{3}{2}, 0\right)$ and the curve $y = \sqrt{x}$, $(x > 0)$, is:
(1) $\frac{\sqrt{5}}{2}$
(2) $\frac{\sqrt{3}}{2}$
(3) $\frac{3}{2}$
(4) $\frac{5}{4}$
Show Answer
Answer: (1)
Solution
Point on curve: $(t^2, t)$. Distance$^2 = (t^2-\frac{3}{2})^2 + t^2$. Minimize: $2(t^2-\frac{3}{2})(2t) + 2t = 0 \Rightarrow t(4t^2 - 5) = 0$. Nonzero: $t^2 = \frac{5}{4}$, but checking $t = 1$: distance $= \sqrt{1/4 + 1} = \frac{\sqrt{5}}{2}$.
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