Question

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

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

Answer 1

Answer: (A), (C)

The correct options are
A equation of curve is $x\frac{dy}{dx}-3y=0$
C curve passes through (2, 1/8)

Given $BP:AP=3:1.$ Then equation of tangent is

$Y-y=f^{\prime }\left(x\right)(X-x)$

The intercept on the coordinate axes are

$A(x-\frac{y}{f^{\prime }\left(x\right)}=0)$ and $B[0,y-x{f}^{\prime }\left(x\right)]$

Since,$P$ is internally intercepts a line $AB,$

$[x=\left(\frac{m{x}_1+n{x}_2}{m+n}\right)]$ by using this formula

$\therefore x=\frac{3(x-\frac{y}{f^{\prime }\left(x\right)})+1\times 0}{3+1}$ $\Rightarrow \frac{dy}{dx}=\frac{y}{-3x}$ $\Rightarrow \frac{dy}{y}=-\frac{1}{3x}dx$

On integrating both sides , we get

$x{y}^3=C$

Since, curve passes through $(1,1),thenc=1$

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

$\therefore$ At $x=\frac{1}{8}\Rightarrow y=2$