Key Concepts and Formulas

• Equation of a Tangent: The equation of the tangent to the curve $y = f(x)$ at the point $(x_1, y_1)$ is given by $y - y_1 = f'(x_1)(x - x_1)$.
• Section Formula: If a point $P(x_1, y_1)$ divides the line segment joining $A(x_A, y_A)$ and $B(x_B, y_B)$ in the ratio $m:n$, then $x_1 = \frac{nx_A + mx_B}{m+n}$ and $y_1 = \frac{ny_A + my_B}{m+n}$.
• Variable Separable Differential Equation: A differential equation of the form $\frac{dy}{dx} = g(x)h(y)$ can be solved by separating variables and integrating: $\int \frac{dy}{h(y)} = \int g(x) dx$.

Step-by-Step Solution

Step 1: Write the equation of the tangent line at point P

Let $P(x_1, y_1)$ be a point on the curve $y = f(x)$. The slope of the tangent at $P$ is $f'(x_1) = \frac{dy}{dx}\Big|_{x=x_1}$. Therefore, the equation of the tangent line at $P$ is:

$y - y_1 = f'(x_1)(x - x_1)$

Step 2: Find the coordinates of the x and y intercepts (points A and B)

We need to find the points where the tangent line intersects the x-axis (point A) and the y-axis (point B).

• Finding point A (x-intercept): The y-coordinate of A is 0. Substitute $y = 0$ into the tangent equation:

  $0 - y_1 = f'(x_1)(x_A - x_1)$ $x_A = x_1 - \frac{y_1}{f'(x_1)}$

  Therefore, the coordinates of A are $A\left(x_1 - \frac{y_1}{f'(x_1)}, 0\right)$.

• Finding point B (y-intercept): The x-coordinate of B is 0. Substitute $x = 0$ into the tangent equation:

  $y_B - y_1 = f'(x_1)(0 - x_1)$ $y_B = y_1 - x_1f'(x_1)$

  Therefore, the coordinates of B are $B\left(0, y_1 - x_1f'(x_1)\right)$.

Step 3: Apply the section formula using the given ratio

We are given that $AP:BP = 1:3$. This means that P divides the line segment AB internally in the ratio 1:3. We can use the section formula to relate the coordinates of A, B, and P.

• Using the x-coordinate:

  $x_1 = \frac{3x_A + 1x_B}{1+3} = \frac{3\left(x_1 - \frac{y_1}{f'(x_1)}\right) + 1(0)}{4}$ $4x_1 = 3x_1 - \frac{3y_1}{f'(x_1)}$ $x_1 = -\frac{3y_1}{f'(x_1)}$ $f'(x_1) = -\frac{3y_1}{x_1}$

• Using the y-coordinate:

  $y_1 = \frac{3y_A + 1y_B}{1+3} = \frac{3(0) + 1(y_1 - x_1f'(x_1))}{4}$ $4y_1 = y_1 - x_1f'(x_1)$ $3y_1 = -x_1f'(x_1)$ $f'(x_1) = -\frac{3y_1}{x_1}$

Both the x and y coordinate equations give the same result: $f'(x_1) = -\frac{3y_1}{x_1}$.

Step 4: Form and solve the differential equation

We can rewrite the equation $f'(x_1) = -\frac{3y_1}{x_1}$ as a differential equation in terms of x and y:

$\frac{dy}{dx} = -\frac{3y}{x}$

This is a separable differential equation. Separate the variables and integrate:

$\frac{dy}{y} = -3\frac{dx}{x}$ $\int \frac{dy}{y} = -3\int \frac{dx}{x}$ $\ln|y| = -3\ln|x| + C$ $\ln|y| = \ln|x^{-3}| + C$ Let $C = \ln|K|$, where K is a constant. Then: $\ln|y| = \ln|Kx^{-3}|$ $y = Kx^{-3} = \frac{K}{x^3}$

Step 5: Apply the initial condition to determine the constant K

We are given that $f(1) = 1$. This means when $x=1$, $y=1$. Substitute these values into the equation:

$1 = \frac{K}{1^3}$ $K = 1$

Thus, the equation of the curve is: $y = \frac{1}{x^3}$

Step 6: Check the given options to find which point lies on the curve

Substitute the x-coordinate of each point into the equation $y = \frac{1}{x^3}$ and see if the resulting y-coordinate matches the given y-coordinate.

• (A) $\left( {{1 \over 3},24} \right)$ If $x = \frac{1}{3}$, then $y = \frac{1}{(1/3)^3} = 27$. This does not match 24.

• (B) $\left( {{1 \over 2},4} \right)$ If $x = \frac{1}{2}$, then $y = \frac{1}{(1/2)^3} = 8$. This does not match 4.

• (C) $\left( {2,{1 \over 8}} \right)$ If $x = 2$, then $y = \frac{1}{2^3} = \frac{1}{8}$. This matches the given y-coordinate.

• (D) $\left( {3,{1 \over 28}} \right)$ If $x = 3$, then $y = \frac{1}{3^3} = \frac{1}{27}$. This does not match $\frac{1}{28}$.

Therefore, the curve passes through the point $\left(2, \frac{1}{8}\right)$.

Common Mistakes & Tips

• Be careful with the signs when applying the section formula and solving for the intercepts.
• Don't forget the constant of integration when solving the differential equation.
• Double-check your calculations, especially when dealing with fractions and exponents.

Summary

We found the equation of the tangent line, used the section formula to relate the coordinates of the points, formed and solved a separable differential equation, and applied the initial condition to find the particular solution. By checking the given options, we found that the curve passes through the point $\left(2, \frac{1}{8}\right)$.

The final answer is \boxed{$\left(2, \frac{1}{8}\right)$}, which corresponds to option (C).