SATHEE: Straight Line and Pair of Straight Lines 1 Question 15

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) $\frac{5}{4}$

(d) $\frac{\sqrt{5}}{2}$

(2019 Main, 10 Jan I)

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Solution:

Let $P (x_{1}, y_{1})$ be any point on the curve $y = \sqrt{x}$ .

Clearly, $y_{1} = \sqrt{x_{1}} \Rightarrow x_{1} = y_{1}^{2} [∵ (x_{1}, y_{1})$ lies on $y = \sqrt{x}]$

$∴$ The point is $P (y_{1}^{2}, y_{1})$

Now, let the given point be $A \frac{3}{2}, 0$ , then

$\begin{aligned} P A & = \sqrt{y_{1}^{2} - {\frac{3}{2}}^{2} + y_{1}^{2}} \\ = \sqrt{y_{1}^{4} - 3 y_{1}^{2} + \frac{9}{4} + y_{1}^{2}} \\ = \sqrt{y_{1}^{4} - 2 y_{1}^{2} + \frac{9}{4}} \\ = \sqrt{{(y_{1}^{2} - 1)}^{2} + \frac{5}{4}} \end{aligned}$

Clearly, $P A$ will be least when

$\begin{aligned} y_{1}^{2} - 1 & = 0 \\ \Rightarrow & P A_{min} = \sqrt{0 + \frac{5}{4}} = \frac{\sqrt{5}}{2} \end{aligned}$

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Straight Line and Pair of Straight Lines 1 Question 15

15. The shortest distance between the point 32,0 and the curve y=x,(x>0), is

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15. The shortest distance between the point $\frac{3}{2}, 0$ and the curve $y = \sqrt{x}, (x > 0)$ , is