Let the tangent at any point P(x, y) on a cur | vedclass

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(A) Let the tangent at $P(x, y)$ intersect the $x$-axis at $A(\alpha, 0)$ and the $y$-axis at $B(0, \beta)$.The equation of the tangent is $Y - y = \f

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$4$

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(A) Let the tangent at $P(x, y)$ intersect the $x$-axis at $A(\alpha, 0)$ and the $y$-axis at $B(0, \beta)$.The equation of the tangent is $Y - y = \frac{dy}{dx}(X - x)$.For $A$,$Y = 0 \implies -y = \frac{dy}{dx}(\alpha - x) \implies \alpha = x - y \frac{dx}{dy}$.For $B$,$X = 0 \implies Y - y = \frac{dy}{dx}(-x) \implies Y = y - x \frac{dy}{dx}$.Given $PA: PB = 1: k$,by section formula,$x = \frac{k \cdot \alpha + 1 \cdot 0}{k + 1} = \frac{k \alpha}{k + 1} \implies \alpha = \frac{k + 1}{k} x$.Substituting $\alpha$: $\frac{k + 1}{k} x = x - y \frac{dx}{dy} \implies \frac{x}{k} = -y \frac{dx}{dy} \implies \frac{dy}{dx} = -\frac{ky}{x}$.Integrating: $\int \frac{dy}{y} = -k \int \frac{dx}{x} \implies \ln y = -k \ln x + C \implies y x^k = C$.Passing through $(1, 1) \implies C = 1$. Passing through $(\frac{1}{10}, 100) \implies 100 \cdot (\frac{1}{10})^k = 1 \implies 10^2 \cdot 10^{-k} = 10^0 \implies 2 - k = 0 \implies k = 2$.The differential equation is $e^{\frac{dy}{dx}} = 2x + 1 \implies \frac{dy}{dx} = \ln(2x + 1)$.Integrating: $y = \int \ln(2x + 1) dx = \frac{1}{2} (2x + 1) \ln(2x + 1) - x + C$.Using $y(0) = 2$: $2 = \frac{1}{2}(1)(0) - 0 + C \implies C = 2$.So,$y(x) = \frac{2x + 1}{2} \ln(2x + 1) - x + 2$.$y(1) = \frac{3}{2} \ln 3 - 1 + 2 = \frac{3}{2} \ln 3 + 1$.$4y(1) - 5 \ln 3 = 4(\frac{3}{2} \ln 3 + 1) - 5 \ln 3 = 6 \ln 3 + 4 - 5 \ln 3 = \ln 3 + 4$.

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