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{d y}{d x}=-\frac{\alpha y}{x} $
$ \frac{d y}{y}=-\frac{\alpha}{x} d x $
$\Rightarrow \frac{d y}{y}+\frac{\alpha}{x} d x=0$
$ \Rightarrow \ell n y+\alpha \ell n x=\ell n c$
$ \Rightarrow yx^\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 $y =2 x ^{-1 / 3}$
At $x=1 / 8, y(1 / 8)=2\left(\frac{1}{8}\right)^{-\frac{1}{3}} \Rightarrow y=4$