Question

A tangent to the curve, $y=f\left(x\right)$ at $P(x,y)$ meets $x-$axis at $A$ and $y-$axis at $B$. If $AP:BP=1:3$ and $f\left(1\right)=1$, then the curve also passes through the point.

• A. $\left(\frac{1}{3},23\right)$
• B. $\left(3,\frac{1}{28}\right)$
• C. $\left(\frac{1}{2},3\right)$
• D. $\left(2,\frac{1}{8}\right)$

Answer 1

The correct option is D $\left(2,\frac{1}{8}\right)$

$\frac{(y-y_1)}{(x-x_1}=f^{\prime }\left(x_1\right)$
$y-y_1=f^{\prime }\left(x_1\right)(x-x_1)$
$y=0,\frac{-y}{f^{\prime }\left(x_1\right)}=x-x_1$
$\Rightarrow x=x_1=\frac{y_1}{f^{\prime }\left(x_1\right)}$
$A=(x_1-\frac{y_1}{f^{\prime }\left(x_1\right)},0)$
$x=0,y-y_1=f^{\prime }\left(x_1\right)\cdot (-x_1)$
$\Rightarrow y=y_1-x_1{f}^{\prime }\left(x_1\right)$
$B=(0,y_1-x_1{f}^{\prime }(x_1\left)\right)$
P divides AB in ratio $1:3$
$\therefore$
$x_1=\frac{3(x_1-\frac{y_1}{f^{\prime }\left(x_1\right)})}{4}$
$y_1=\frac{y_1-x_1{f}^{\prime }\left(x_1\right)}{4}$
$4y_1=y_1{x}_1{f}^{\prime }\left(x_1\right)$
$f^{\prime }\left(x_1\right)=\frac{-3y_1}{x_1}$
$f^{\prime }\left(x\right)=\frac{-3y}{x}$
$\frac{dy}{dx}=\frac{-3y}{x}$
$\frac{dy}{y}=\frac{-3dx}{x}$
$lny=-3lnx+c$
$y=k{x}^{-3}$
$y\left(1\right)=1$
$\Rightarrow k=1$
$y=\frac{1}{x^3}$.

Option $D$ satisfies the above function.