Question

A tangent to the curve, $y = f\left( x \right)$ at $P\left( {x,y} \right)$ meets the x-axis at A and y-axis at B. If $AP:BP = 1:3$ and $f\left( 1 \right) = 1$, then the curve also passes through the point.
A. $\left( {\dfrac{1}{3},23} \right)$
B. $\left( {3,\dfrac{1}{8}} \right)$
C. $\left( {\dfrac{1}{2},3} \right)$
D. $\left( {2,\dfrac{1}{8}} \right)$

Answer 1

Hint: To solve the question, at first we have to find out the equation of the tangent to the curve $P\left( {x,y} \right)$. Then we will find out the x-intercept and y-intercept of the tangent by substituting $Y = 0$ and $X = 0$ in the tangent equation. Hence we can get the coordinates of A and B. Since $P\left( {x,y} \right)$ divides AB with a ratio 1:3, by using ratio formula, we must find out the coordinate of P and then equate its abscissa and ordinate with x and y respectively and obtain the equations. Finally solving the equations we can get the equation of the curve and the correct coordinate in the options must satisfy the equation of the curve.

Complete step-by-step solution:
Given a point on the curve where the tangent drawn is $P\left( {x,y} \right)$.
We know that the equation of a tangent at the point $P\left( {x,y} \right)$ to a curve $Y = f\left( x \right)$ is given by,
$ \Rightarrow \left( {Y - y} \right) = f'\left( x \right)\left( {X - x} \right)$...............….. (1)
Where $f'\left( x \right)$ is the slope of the curve at $P\left( {x,y} \right)$.
The x-intercept of the tangent can be found out by substituting $Y = 0$ in equation (1),
$ \Rightarrow \left( {0 - y} \right) = f'\left( x \right)\left( {X - x} \right)$
Divide both sides by $f'\left( x \right)$,
$ \Rightarrow X - x = - \dfrac{y}{{f'\left( x \right)}}$
Move $x$ on the right side,
$ \Rightarrow X = x - \dfrac{y}{{f'\left( x \right)}}$................….. (2)
The y-intercept of the tangent can be found out by substituting $X = 0$ in equation (1),
$ \Rightarrow \left( {Y - y} \right) = f'\left( x \right)\left( {0 - x} \right)$
Simplify the terms,
$ \Rightarrow Y - y = - xf'\left( x \right)$
Move $y$ on the right side,
$ \Rightarrow Y = y - xf'\left( x \right)$.................….. (3)
Since the tangent meets the x-axis at A and the y-axis at B then the coordinate of A is \[\left( {X,0} \right) = \left( {x - \dfrac{y}{{f'\left( x \right)}},0} \right)\] and the coordinate of B is $\left( {0,Y} \right) = \left( {0,y - xf'\left( x \right)} \right)$.
We know the section formulae that the coordinate of a point that divides the line segment joining the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ in the ratio $m:n$ is $\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$.
Since it is given that $AP:PB = 1:3$ that means $P\left( {x,y} \right)$ divide \[A\left( {x - \dfrac{y}{{f'\left( x \right)}},0} \right)\] and $B\left( {0,y - xf'\left( x \right)} \right)$ in a ratio $1:3$.
So, applying the section formula the coordinate of P is given by
$ \Rightarrow \left( {\dfrac{{1 \times 0 + 3\left( {x - \dfrac{y}{{f'\left( x \right)}}} \right)}}{{1 + 3}},\dfrac{{1\left( {y - xf'\left( x \right) + 3 \times 0} \right)}}{{1 + 3}}} \right) = \left( {\dfrac{{3\left( {x - \dfrac{y}{{f'\left( x \right)}}} \right)}}{4},\dfrac{{y - xf'\left( x \right)}}{4}} \right)$
But given that the coordinate of P is $\left( {x,y} \right)$. Then, equating the respective abscissa and ordinates we will get
$ \Rightarrow \dfrac{{3\left( {x - \dfrac{y}{{f'\left( x \right)}}} \right)}}{4} = x$
Divide both sides by $\dfrac{4}{3}$,
$ \Rightarrow x - \dfrac{y}{{f'\left( x \right)}} = \dfrac{{4x}}{3}$
Move $x$ component on one side,
$ \Rightarrow \dfrac{y}{{f'\left( x \right)}} = x - \dfrac{{4x}}{3}$
Simplify the terms,
$ \Rightarrow \dfrac{y}{{f'\left( x \right)}} = - \dfrac{x}{3}$
Cross multiply the terms,
$ \Rightarrow - xf'\left( x \right) = 3y$
Divide both sides by $ - x$,
\[ \Rightarrow f'\left( x \right) = - \dfrac{{3y}}{x}\]
Convert $f'\left( x \right)$ in $\dfrac{{dy}}{{dx}}$,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{3y}}{x}\]
Simplify the terms,
\[ \Rightarrow \dfrac{{dy}}{y} = - 3\dfrac{{dx}}{x}\]
Now integrate the differential equation to get the solution,
$ \Rightarrow \int {\dfrac{{dy}}{y} = - 3\int {\dfrac{{dx}}{x}} } $
Apply integration on both sides,
\[ \Rightarrow \ln y = - 3\ln x + c\]....................….. (4)
But it is given that $f\left( 1 \right) = 1$ that means $x = 1,y = 1$. Substitute the values in equation (4),
\[ \Rightarrow \ln 1 = - 3\ln 1 + c\]
Simplify the terms,
$ \Rightarrow c = 0$
Substitute the value in equation (4),
\[ \Rightarrow \ln y = - 3\ln x\]
Simplify the terms,
\[ \Rightarrow \ln y = \ln {x^{ - 3}}\]
Cancel out ln from both sides,
$\therefore y = \dfrac{1}{{{x^3}}}$
When we substitute the value from given options, only option (C) that is $\left( {2,\dfrac{1}{8}} \right)$ satisfies the above equation.

Hence, option (D) is the correct answer.

Note: If a point having coordinate $\left( {{x_1},{y_1}} \right)$ satisfies an equation $y = f\left( x \right)$ then ${y_1} = f\left( {{x_1}} \right)$ that means the abscissa and ordinate of the point replaces the value of $x$ and $y$ from the equation.