Question

A tangent to the curve $y=f\left(x\right)$ at$P(x,y)$ cuts the x-axis and the y-axis at A and B, respectively, such that$BP:AP=3:1.$ If$f\left(1\right)=1$ then :

• A. the equation of the curve is $x\frac{dy}{dx}+3y=0$
• B. the curve passes through $(2,\frac{1}{8})$
• C. the equation of the curve is $x\frac{dy}{dx}-3y=0$
• D. the normal at$(1,1)$ is $x+3y=4$

Answer 1

Answer: (A), (B)

the equation of the curve is $x\frac{dy}{dx}+3y=0$
the curve passes through $(2,\frac{1}{8})$

Equation of tangent is $Y-y=\frac{dy}{dx}(X-x)$
Given $\frac{BP}{AP}=\frac{3}{1}$ so that
$\Rightarrow \frac{dx}{x}=-\frac{dy}{3y}\Rightarrow x\frac{dy}{dx}+3y=0$
$\Rightarrow \tan x^3=-\tan cy\Rightarrow \frac{1}{x^3}=cy$
Given $f\left(1\right)=1\Rightarrow c=1$
$\therefore y=\frac{1}{x^2}$
and point $(2,\frac{1}{8})$ satisfy this.