A tangent to the curve y = f(x) at P(x, y) me | vedclass

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(C) Let the tangent at $P(x, y)$ be $Y - y = f'(x)(X - x)$.For $A$ (where $Y=0$),$X = x - \frac{y}{f'(x)}$. So $A = \left( x - \frac{y}{f'(x)}, 0 \rig

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$\left( 2, \frac{1}{8} \right)$

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(C) Let the tangent at $P(x, y)$ be $Y - y = f'(x)(X - x)$.For $A$ (where $Y=0$),$X = x - \frac{y}{f'(x)}$. So $A = \left( x - \frac{y}{f'(x)}, 0 \right)$.For $B$ (where $X=0$),$Y = y - x f'(x)$. So $B = (0, y - x f'(x))$.Given $AP : BP = 1 : 3$,by section formula for $P(x, y)$ dividing $AB$ in ratio $1:3$:$x = \frac{1 \cdot 0 + 3 \cdot (x - y/f'(x))}{1 + 3} \implies 4x = 3x - \frac{3y}{f'(x)} \implies x = -\frac{3y}{f'(x)}$.Thus,$\frac{dy}{dx} = -\frac{3y}{x}$.Separating variables: $\int \frac{dy}{y} = -3 \int \frac{dx}{x} \implies \ln|y| = -3 \ln|x| + C \implies y = \frac{k}{x^3}$.Given $f(1) = 1$,we get $1 = \frac{k}{1^3} \implies k = 1$. So $y = \frac{1}{x^3}$.Checking the options,for $x=2$,$y = \frac{1}{2^3} = \frac{1}{8}$.Thus,the curve passes through $\left( 2, \frac{1}{8} \right)$.

$A$ tangent to the curve $y = f(x)$ at $P(x, y)$ meets the $x-$axis at $A$ and the $y-$axis at $B$. If $AP : BP = 1 : 3$ and $f(1) = 1$,then the curve also passes through the point:

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