Question

Tangent is drawn at any point $P(x,y)$ on a curve, which passes through $(1,1)$. The tangent cuts $X$-axis and $Y$-axis at $A$ and $B$ respectively. If $AP:BP=3:1$, then

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

Answer 1

Answer: (C)

$\frac{PA}{PB}=\frac{3}{1}$
Equation of tangent $AB$ is $Y-y=\frac{dy}{dx}(X-x)$
$A\left(\frac{x{y}^{\prime }-y}{y^{\prime }},0\right)$ and $B\left(0,y-x{y}^{\prime }\right)$
Using section formula : $y=\frac{1\times 0+3\times \left(y-x{y}^{\prime }\right)}{4}$
$\Rightarrow 4y=3y-3x{y}^{\prime }$
$\Rightarrow 3x{y}^{\prime }=-y$
$\Rightarrow \frac{3xdy}{dx}+y=0$
$\frac{3dy}{y}+\frac{dx}{x}=0$
$x{y}^3=1$
$\frac{dy}{d{x}_{(1,1)}}=-\frac{1}{3}$
Slope of Normal $=3$
Equation of Normal
$\Rightarrow y-1=3(x-1)$
$\Rightarrow y-3x+2=0$