Straight Line and Pair of Straight Lines 1 Question 15
15. The shortest distance between the point $\frac{3}{2}, 0$ and the curve $y=\sqrt{x},(x>0)$, is
(a) $\frac{3}{2}$
(b) $\frac{5}{4}$
(c) $\frac{\sqrt{3}}{2}$
(d) $\frac{\sqrt{5}}{2}$
(2019 Main, 10 Jan I)
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Solution:
- Let $P\left(x _1, y _1\right)$ be any point on the curve $y=\sqrt{x}$.
Clearly, $y _1=\sqrt{x _1} \Rightarrow x _1=y _1^{2}\left[\because\left(x _1, y _1\right)\right.$ lies on $\left.y=\sqrt{x}\right]$
$\therefore$ The point is $P\left(y _1^{2}, y _1\right)$
Now, let the given point be $A \frac{3}{2}, 0$, then
$$ \begin{aligned} P A & =\sqrt{y _1^{2}-\frac{3}{2}^{2}+y _1^{2}} \\ & =\sqrt{y _1^{4}-3 y _1^{2}+\frac{9}{4}+y _1^{2}} \\ & =\sqrt{y _1^{4}-2 y _1^{2}+\frac{9}{4}} \\ & =\sqrt{\left(y _1^{2}-1\right)^{2}+\frac{5}{4}} \end{aligned} $$
Clearly, $P A$ will be least when
$$ \begin{aligned} y _1^{2}-1 & =0 \\ \Rightarrow & P A _{\min }=\sqrt{0+\frac{5}{4}}=\frac{\sqrt{5}}{2} \end{aligned} $$
