Question

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 point $(1,2)$ and $(8,1)$, then $\left|y\left(\frac{1}{8}\right)\right|$ is equal to

• A. $2{\log }_{e}2$
• B. 4
• C. 1
• D. $4{\log }_{e}2$

Answer 1

Answer: (B)

$\frac{dy}{dx}=-\frac{\alpha y}{x}$
$\frac{dy}{y}=-\frac{\alpha }{x}dx$
$\Rightarrow \frac{dy}{y}+\frac{\alpha }{x}dx=0$
$\Rightarrow \ell ny+\alpha \ell nx=\ell nc$
$\Rightarrow y{x}^{\alpha }=c$
For $(1,2)\Rightarrow 2.1^{\alpha }=c\Rightarrow c=2$
For $(8,1)\Rightarrow 1.8^{\alpha }=2\Rightarrow \alpha =\frac{1}{3}$
$\therefore$ curve is $y=2x^{-1/3}$
At $x=1/8,y(1/8)=2{\left(\frac{1}{8}\right)}^{-\frac{1}{3}}\Rightarrow y=4$