Question

A tangent at point $P(x,y)$ on the curve $y=f\left(x\right)$ intersects to the axes at A and B points respectively such that $AP:BP\equiv 1:3$. Given that $f\left(1\right)=1$, then

• A. Normal at $(1,1)$ is $x+3y=4$
• B. Equation of curve is $3y+x{y}^{\prime }=0$
• C. Curve passes through $(2,1/8)$
• D. Equation of curve is $x{y}^{\prime }-3y=0$

Answer 1

Answer: (C), (D)

The correct options are
C Curve passes through $(2,1/8)$
D Equation of curve is $x{y}^{\prime }-3y=0$

Eqution of the tangent is

$Y-y=\frac{dy}{dx}(X-x)$

Given

$\frac{BP}{AP}=\frac{3}{1}$

$\Rightarrow \frac{dx}{x}=\frac{dy}{3y}\Rightarrow x\times \frac{dy}{dx}+3y=0$

$\Rightarrow lnx=-\frac{1}{3}\times lny-lnc=ln{x}^3=-\left(lncy\right)$

$\Rightarrow \frac{1}{x^3}=cy,$ Given $f\left(1\right)=1\Rightarrow c=1$

$\therefore y=\frac{1}{x^3}$