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 $\overline{AP}:\overline{BP}=1:3$ and $f\left(1\right)=1$ then the curve also passes through the point.

• A. $(\frac{1}{2},4)$
• B. $(\frac{1}{3},24)$
• C. $(2,\frac{1}{8})$
• D. $(3,\frac{1}{28})$

Answer 1

The correct option is C $(2,\frac{1}{8})$

$\frac{(y-y_1)}{(x-x_1)}=f^{\prime }\left(x_1\right)$

$y-y_1=f^{\prime }\left(x_1\right)(x-x_1)$

$y=0,\frac{-y}{f^{\prime }\left(x_1\right)}=x-x_1$

$\Rightarrow x=x_1=\frac{y_1}{f^{\prime }\left(x_1\right)}$

$A=(x_1-\frac{y_1}{r^{\prime }\left(x_1\right)},0)$

$x=0,y-y_1=f^{\prime }\left(x_1\right).(-x_1)$

$\Rightarrow y=y_1-x_1{f}^{\prime }\left(x_1\right)$

$B=(0,y_1-x_1{f}^{\prime }(x_1\left)\right)$

P divides $AB$ in ration $1:3$

$x_1=\frac{3[x_1-\frac{y_1}{f^{\prime }\left(x_1\right)}]}{4},y_1=\frac{[y_1-x_1{f}^{\prime }(x_1\left)\right]}{4}$

$4y_1=y_1{x}_1{f}^{\prime }\left(x_1\right)$

$f^{\prime }\left(x_1\right)=\frac{-3y_1}{x_1}$

$f^{\prime }\left(x\right)=\left(\frac{-3y}{x}\right)$

$\frac{dy}{dx}=\frac{-3y}{x}$

$\frac{dy}{y}=\frac{-3dx}{x}$

$\ln y=-3\ln x+c$

$y=k{x}^{-3}$

$y\left(1\right)=1$

$\Rightarrow k=1\Rightarrow y=\frac{1}{x^3}$ put $x=2,y=\frac{1}{8}(2,\frac{1}{8})$