A tangent is drawn at any point P(x, y) on a | vedclass

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(A) Let the tangent at $P(x, y)$ be $Y - y = \frac{dy}{dx}(X - x)$.Let $y' = \frac{dy}{dx}$. The tangent is $Y - y = y'(X - x)$.For point $A$ ($X$-axi

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the differential equation of the curve is $3x \frac{dy}{dx} + y = 0$

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(A) Let the tangent at $P(x, y)$ be $Y - y = \frac{dy}{dx}(X - x)$.Let $y' = \frac{dy}{dx}$. The tangent is $Y - y = y'(X - x)$.For point $A$ ($X$-axis), set $Y = 0$: $-y = y'(X - x) \Rightarrow X = x - \frac{y}{y'}$. So, $A = \left(x - \frac{y}{y'}, 0\right)$.For point $B$ ($Y$-axis), set $X = 0$: $Y - y = y'(-x) \Rightarrow Y = y - xy'$. So, $B = (0, y - xy')$.Given $AP:BP = 3:1$. Using the section formula for point $P(x, y)$ dividing $AB$ in ratio $3:1$:$x = \frac{1 \cdot (x - y/y') + 3 \cdot 0}{3 + 1} = \frac{x - y/y'}{4} \Rightarrow 4x = x - \frac{y}{y'} \Rightarrow 3x = -\frac{y}{y'} \Rightarrow 3xy' = -y \Rightarrow 3x \frac{dy}{dx} + y = 0$.This matches option $A$.Solving the differential equation: $\frac{3 dy}{y} + \frac{dx}{x} = 0 \Rightarrow 3 \ln|y| + \ln|x| = C \Rightarrow \ln|xy^3| = C \Rightarrow xy^3 = k$.Since it passes through $(1, 1)$, $1(1)^3 = k \Rightarrow k = 1$. The curve is $xy^3 = 1$.Check option $C$: If $x = 1/8$, $y^3 = 8 \Rightarrow y = 2$. So, the curve passes through $(1/8, 2)$.Check option $D$: At $(1, 1)$, $y' = -\frac{y}{3x} = -\frac{1}{3}$. Slope of normal $= -\frac{1}{y'} = 3$.Equation of normal: $y - 1 = 3(x - 1) \Rightarrow y - 1 = 3x - 3 \Rightarrow 3x - y = 2$. Option $D$ is incorrect.

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

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