    Question

# A tangent drawn to the curve $y=f\left(x\right)$ at $P\left(x,y\right)$ 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 $x\left(\frac{dy}{dx}\right)-3y=0$

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B

Normal at $\left(1,1\right)$ is $x+3y=4$

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C

Curve passes through$\left(2,\frac{1}{8}\right)$

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D

Equation of the curve is $x\left(\frac{dy}{dx}\right)+3y=0$

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E

$3$ and $4$

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Solution

## The correct option is E $3$ and $4$Explanation for the correct option: Step 1. Forming the intercept axesGiven $BP:AP=3:1$ Then equation of tangent is $Y-y=f\left(x\right)\left(X-x\right)$The intercept on the coordinate axes are$A\left(x-\frac{y}{f\left(x\right)}=0\right),B\left[0,y-xf\left(x\right)\right)$Since, P is internally intercepts a line AB,$x=\frac{3\left(\frac{y}{f\left(x\right)}\right)+1×0}{3+1}\mathbf{}\mathbf{\left[}\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{\right]}\mathbf{}\phantom{\rule{0ex}{0ex}}$$⇒\frac{dy}{dx}=\frac{y}{-3x}$$⇒\frac{dy}{y}=-\frac{1}{3x}dx$Step 2. On integrating both sides, we get$x{y}^{3}=c$Since, Curve passes through $\left(1,1\right)$ then $c=1$$\therefore x{y}^{3}=1$At $x=\frac{1}{8}$$⇒$$y=2$$\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.  Suggest Corrections  0      Similar questions  Explore more