"Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1,2)$ and $(8,1)$, then $\left|y\left(\frac{1}{8}\right)\right|$ is equal to"

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${{dy} \over {dx}} \propto {{ - y} \over x}$

$${{dy} \over {dx}} = {{ - ky} \over x} \Rightarrow \int {{{dy} \over y} = - K\int {{{dx} \over x}} } $$

$\ln |y| = - K\ln |x| + C$

If the above equation satisfy (1, 2) and (8, 1)

$\ln 2 = - K \times 0 + C \Rightarrow C = \ln 2$

$\ln 1 = - K\ln 8 + \ln 2 \Rightarrow K = {1 \over 3}$

So, at $x = {1 \over 8}$

$\ln |y| = - {1 \over 3}\ln \left( {{1 \over 8}} \right) + \ln 2 = 2\ln 2$

$|y| = 4$

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Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1,2)$ and $(8,1)$, then $\left|y\left(\frac{1}{8}\right)\right|$ is equal to

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Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1,2)$ and $(8,1)$, then $\left|y\left(\frac{1}{8}\right)\right|$ is equal to

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