The tangent to a non-linear curve y = f(x) at | vedclass

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(A) Let the point $P$ be $(x, y)$. The equation of the tangent at $P$ is $Y - y = f'(x)(X - x)$.For point $A$ (where $Y=0$),$X = x - \frac{y}{f'(x)}$,

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$y = \frac{6}{x}$

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(A) Let the point $P$ be $(x, y)$. The equation of the tangent at $P$ is $Y - y = f'(x)(X - x)$.For point $A$ (where $Y=0$),$X = x - \frac{y}{f'(x)}$,so $A = (x - \frac{y}{f'(x)}, 0)$.For point $B$ (where $X=0$),$Y = y - x f'(x)$,so $B = (0, y - x f'(x))$.The normal at $P$ has slope $-\frac{1}{f'(x)}$. Its equation is $Y - y = -\frac{1}{f'(x)}(X - x)$.For point $C$ (where $X=0$),$Y = y + \frac{x}{f'(x)}$,so $C = (0, y + \frac{x}{f'(x)})$.Given $AC = BC$,which implies $AC^2 = BC^2$. Since $A$ is on the $x$-axis and $B, C$ are on the $y$-axis,the distance formula gives:$(x - \frac{y}{f'(x)})^2 + (y + \frac{x}{f'(x)})^2 = (y - x f'(x) - (y + \frac{x}{f'(x)}))^2$$(x - \frac{y}{f'(x)})^2 + (y + \frac{x}{f'(x)})^2 = (-x f'(x) - \frac{x}{f'(x)})^2$$x^2 - \frac{2xy}{f'(x)} + \frac{y^2}{(f'(x))^2} + y^2 + \frac{2xy}{f'(x)} + \frac{x^2}{(f'(x))^2} = x^2 (f'(x) + \frac{1}{f'(x)})^2$$x^2 + y^2 + \frac{x^2 + y^2}{(f'(x))^2} = x^2 (f'(x)^2 + 2 + \frac{1}{f'(x)^2})$$(x^2 + y^2)(1 + \frac{1}{(f'(x))^2}) = x^2 (f'(x)^2 + 2 + \frac{1}{f'(x)^2})$This simplifies to $f'(x) = -\frac{y}{x}$.Integrating $\frac{dy}{y} = -\frac{dx}{x}$ gives $\ln y = -\ln x + \ln c$,so $xy = c$.Using $f(2) = 3$,we get $2 	imes 3 = c$,so $c = 6$.The equation is $xy = 6$ or $y = \frac{6}{x}$.

The tangent to a non-linear curve $y = f(x)$ at any point $P(x, y)$ intersects the $x$-axis and $y$-axis at $A$ and $B$ respectively. If the normal to the curve $y = f(x)$ at $P$ intersects the $y$-axis at $C$ such that $AC = BC$,and $f(2) = 3$,then the equation of the curve is:

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