Question

A tangent drawn to the curve $y=f\left(x\right)$ at $P\left(x,y\right)$
cuts the x and y axes at A and B, respectively, such that $AP:PB=1:3$. If $f\left(1\right)=1$ then the curve passes through $\left(k,\frac{1}{8}\right)$ where $k$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Equation of tangent at $P(x,y)$ is

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

It meets x - axis at $A$

$A=(x-\frac{dx}{dy}y,0)$

and y -axis at $B$

$B=(0,y-\frac{dy}{dt})$

And also given $\frac{AP}{PB}=1:3$

Using section formula

$[\frac{m{x}_1+m_2{x}_2}{m_{1}m+m_2},\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}]$

$\frac{3}{4}(x-\frac{dx}{dy}y)v=\frac{1}{4}(y-x\frac{dy}{dx})=(x,y)$

$\frac{3}{4}(x-y\frac{dx}{dy})=x$; $\frac{1}{4}(y-x\frac{dy}{x})=y$

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

$\log \left(x^{3}y\right)=$ const (on integrative)

Given $f\left(1\right)=1$, so here $const=1$

$x^{3}y=1$ (k, 1/8)

$k^3=8$

$k=2$