A tangent to the curve, y = f(x) at P(x, y) meets x-axis at A and 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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Official solution
We have
\[{{(y - {y_2})} \over {(x - {x_1})}} = f'({x_1})\]
\[ \Rightarrow y - {y_1} = f'({x_1})(x - {x_1})\]
\[\bullet\] When y = 0: \[{{ - {y_1}} \over {f'({x_1})}} = x - {x_1}\]
\[ \Rightarrow x = {x_1} - {{{y_1}} \over {f'({x_1})}}\]
Therefore, point A is \[A\left( {{x_1} - {{{y_1}} \over {f'({x_1})}},0} \right)\]
\[\bullet\] When x = 0: \[y - {y_1} = f'(x)\,.\,( - {x_1})\]
\[ \Rightarrow y = {y_1} - {x_1}f'({x_1})\]
Therefore, point B is B(0,y1−x1f′(x1))
Point P divides AB in the ratio 1 : 3.
\[{x_1} = {{\left[ {3\left( {{x_1} - {{{y_1}} \over {f'({x_1})}}} \right)} \right]} \over 4}\]
\[{y_1} = {{{y_1} - {x_1}f'({x_1})} \over 4}\]
Therefore, 4y1=y1−x1f′(x1)
\[ \Rightarrow f'({x_1}) = {{ - 3{y_1}} \over {{x_1}}} \Rightarrow f'(x) = {{ - 3y} \over x}\]
Now, \[{{dy} \over {dx}} = {{ - 3y} \over x} \Rightarrow {{dy} \over y} = {{ - 3dx} \over x}\]
On integrating, we get
\[\ln y = - 3\ln x + C \Rightarrow y = k{x^{ - 3}}\]
\[y(1) = 1 \Rightarrow k = 1\]
\[y = {1 \over {{x^3}}}\]
When we substitute the values from the given options, only option (c) satisfies the above equation.