Let a smooth curve y = f(x) be such that the slope of the tangent...

Given, $\dfrac{dy}{dx} ∝ \dfrac{- y}{x}$

$⇒ \dfrac{dy}{dx} = \dfrac{- Ky}{x}$ (where $K$ is proportionality constant)

Now integrating both side,

$⇒ ∫ \dfrac{dy}{y} = - K ∫ \dfrac{dx}{x}$

$⇒ \ln \left|y\right| = - Kln\left|x\right| + C$

If the above equation satisfy \left(1, 2\right)$</p><p>$⇒ \ln 2 = - K \times 0 + C ⇒ C = \ln 2$</p><p>So,$\ln \left|y\right| = - Kln\left|x\right| + \ln 2$</p><p>Now it also passes through \left(8, 1\right)$

$⇒ \ln 1 = - Kln8 + \ln 2 ⇒ K = \dfrac{1}{3}$

So, equation becomes $\ln \left|y\right| = - \dfrac{1}{3}\ln \left|x\right| + \ln 2$

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

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

$⇒ \left|y\right| = 4$