Question

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.
(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 tangent to the curve \[P(x,y)\]. 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 \[P(x,y)\]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 curve.

Complete step-by-step answer:
Given a point on the curve where the tangent drawn is \[P(x,y)\].
We know that the equation of a tangent at point \[P(x,y)\] to a curve \[Y = f(X)\] is given by,
\[\left( {Y - {y_1}} \right) = f'(x)\left( {X - x} \right)\] ……………………………. (1)
Where \[f'(x)\] is the slope of the curve at \[P(x,y)\]
The y-intercept of the of the tangent can be found out by substituting \[Y = 0\] in eq. (1), we will get,
\[\begin{gathered}
   \Rightarrow \left( {0 - y} \right) = f'(x)\left( {X - x} \right) \\
   \Rightarrow X = x - \dfrac{y}{{f'(x)}} \\
\end{gathered} \]
………………………………….. (2)
The y-intercept of the of the tangent can be found out by substituting \[X = 0\] in eq. (1), we will get
\[\begin{gathered}
   \Rightarrow \left( {Y - y} \right) = f'(x)\left( {0 - x} \right) \\
   \Rightarrow Y = y - xf'(x) \\
\end{gathered} \]
……………………………………. (3)
Since the tangent meets x-axis at A and y-axis at B then the coordinate of A is \[\left( {X,0} \right) = \left( {x - \dfrac{y}{{f'(x)}},0} \right)\]and the coordinate of B is\[\left( {0,Y} \right) = \left( {0,y - xf'(x)} \right)\].
We know the section formulae that the coordinate of a point that divides the line segment joining the points \[({x_1},{y_1})\]and \[({x_2},{y_2})\] 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 given that \[AP:BP = 1:3\]that means \[P(x,y)\]divides \[A\left( {x - \dfrac{y}{{f'(x)}},0} \right)\]and \[B\left( {0,y - xf'(x)} \right)\]in a ratio 1:3. Hence applying the section formula the coordinate of P is given by\[\left( {\dfrac{{1 \times 0 + 3\left( {x - \dfrac{y}{{f'(x)}}} \right)}}{{1 + 3}},\dfrac{{1\left( {y - xf'(x)} \right) + 3 \times 0}}{{1 + 3}}} \right) = \left( {\dfrac{{3\left( {x - \dfrac{y}{{f'(x)}}} \right)}}{4},\dfrac{{y - xf'(x)}}{4}} \right)\].
But given that the coordinate of P is\[\left( {x,y} \right)\]. Therefore equating the respective abscissa and ordinates we will get
\[\begin{gathered}
   \Rightarrow \dfrac{{3\left( {x - \dfrac{y}{{f'(x)}}} \right)}}{4} = x \\
   \Rightarrow x = - \dfrac{{3y}}{{f'(x)}} \\
   \Rightarrow f'(x) = - \dfrac{{3y}}{x} \\
\end{gathered} \]
………………………………….. (4)
And
 \[\begin{gathered}
   \Rightarrow \dfrac{{y - xf'(x)}}{4} = y \\
   \Rightarrow f'(x) = - \dfrac{{3y}}{x} \\
\end{gathered} \]
Now we got the differential equation,\[f'(x) = - \dfrac{{3y}}{x}\]. The solution is given by,
\[\begin{gathered}
   \Rightarrow f'(x) = - \dfrac{{3y}}{x} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{3y}}{x} \\
   \Rightarrow \dfrac{{dy}}{y} = - \dfrac{{3dx}}{x} \\
   \Rightarrow \int {\dfrac{{dy}}{y}} = - 3\int {\dfrac{{dx}}{x}} \\
   \Rightarrow \ln y = - 3\ln x + c \\
\end{gathered} \]
………………………………………….. (5)
But given\[f(1) = 1\], that means for \[x = 1,y = 1\].Substituting these values in equation (5) we will get
 \[\begin{gathered}
   \Rightarrow \ln 1 = - 3\ln 1 + c \\
   \Rightarrow c = 0 \\
\end{gathered} \]
………………………………… (6)
Now substituting the value of eq. (6) in eq. (5) we will get
\[\begin{gathered}
   \Rightarrow \ln y = - 3\ln x \\
   \Rightarrow \ln y = \ln \left( {\dfrac{1}{{{x^3}}}} \right) \\
   \Rightarrow y = \dfrac{1}{{{x^3}}} \\
\end{gathered} \]
……………………………….. (7)
When we substitute the value from given options, only option (D) that is \[\left( {2,\dfrac{1}{8}} \right)\] satisfies the above equation.
Hence option (D) is the required answer.

Note: If a point having coordinate \[({x_1},{y_1})\] satisfies an equation \[y = f(x)\]then\[{y_1} = f({x_1})\] that means the abscissa and ordinate of the point replaces x, y and z respectively in the equation.