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 the curve is $x\left(\frac{dy}{dx}\right)-3y=0$
• B. Normal at $(1,1)$ is $x+3y=4$
• C. Curve passes through$\left(2,\frac{1}{8}\right)$
• D. Equation of the curve is $x\left(\frac{dy}{dx}\right)+3y=0$
• E. $3$ and $4$

Answer 1

The correct option is E

$3$ and $4$

Explanation for the correct option:

Step 1. Forming the intercept axes

Given $BP:AP=3:1$

Then equation of tangent is $Y-y=f\left(x\right)(X-x)$

The intercept on the coordinate axes are

$A\left(x-\frac{y}{f\left(x\right)}=0),B[0,y-xf\left(x\right)\right)$

Since, P is internally intercepts a line AB,$x=\frac{3\left({\displaystyle \frac{y}{f\left(x\right)}}\right)+1\times 0}{3+1}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{x}\mathbf{=}\mathbf{\left(}\frac{{\mathbf{mx}}_{\mathbf{1}}\mathbf{+}{\mathbf{nx}}_{\mathbf{2}}}{\mathbf{m}\mathbf{+}\mathbf{n}}\mathbf{\right)}\mathbf{]}\mathbf{}$

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

$\Rightarrow \frac{dy}{y}=-\frac{1}{3x}dx$

Step 2. On integrating both sides, we get

$x{y}^3=c$

Since, Curve passes through $(1,1)$ then $c=1$

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

At $x=\frac{1}{8}$

$\Rightarrow$$y=2$

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

Thus, the curve passes through $\left(2,\frac{1}{8}\right)$ and the equation of curve is $x\left(\frac{dy}{dx}\right)+3y=0$

Hence, Option ‘E’ is Correct.