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 $A P: B P=3: 1$, then

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

Answer 1

Answer: (C)

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