Key Concepts and Formulas

• Derivative as Slope of Tangent: The derivative of a function $y = f(x)$ at a point $(x_0, y_0)$ gives the slope of the tangent line to the curve at that point: $m = \frac{dy}{dx}\Big|_{(x_0, y_0)}$.
• Equation of a Tangent Line: The equation of the tangent line to the curve $y = f(x)$ at the point $(x_0, y_0)$ is given by $y - y_0 = m(x - x_0)$, where $m$ is the slope of the tangent.
• Section Formula: If a point $(x, y)$ divides the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in the ratio $m:n$, then $x = \frac{mx_2 + nx_1}{m+n}$ and $y = \frac{my_2 + ny_1}{m+n}$.

Step-by-Step Solution

Step 1: Find the derivative of the curve.

• Why: We need the derivative to find the slope of the tangent at the point $P(t, t^3)$.
• Calculation: Given $y = x^3$, we differentiate with respect to $x$: $\frac{dy}{dx} = 3x^2$

Step 2: Find the slope of the tangent at point P.

• Why: We need the slope to write the equation of the tangent line.
• Calculation: Evaluate the derivative at $x = t$: $m = \frac{dy}{dx}\Big|_{x=t} = 3t^2$

Step 3: Find the equation of the tangent line at point P.

• Why: We need the equation of the tangent line to find the coordinates of point Q, where the tangent intersects the curve again.
• Calculation: Using the point-slope form of a line, the equation of the tangent at $P(t, t^3)$ is: $y - t^3 = 3t^2(x - t)$ $y = 3t^2x - 3t^3 + t^3$ $y = 3t^2x - 2t^3$

Step 4: Find the coordinates of point Q.

• Why: Point Q is the intersection of the tangent line and the original curve. We need to find its coordinates to use the section formula.
• Calculation: To find the intersection, we set the equation of the tangent equal to the equation of the curve: $x^3 = 3t^2x - 2t^3$ $x^3 - 3t^2x + 2t^3 = 0$ We know that $x = t$ is a solution (since the tangent touches the curve at P). Therefore, $(x - t)$ is a factor. We can factor the cubic equation: $(x - t)(x^2 + tx - 2t^2) = 0$ $(x - t)(x - t)(x + 2t) = 0$ $(x - t)^2(x + 2t) = 0$ The solutions are $x = t$ (which corresponds to point P) and $x = -2t$. Thus, the x-coordinate of point Q is $x_Q = -2t$. The y-coordinate of point Q is $y_Q = (-2t)^3 = -8t^3$. So, $Q = (-2t, -8t^3)$.

Step 5: Apply the section formula to find the coordinates of the point dividing PQ in the ratio 1:2.

• Why: We need to find the ordinate of the point that divides PQ in the ratio 1:2.
• Calculation: Let the point be $R(x, y)$. Using the section formula with $P(t, t^3)$, $Q(-2t, -8t^3)$, and the ratio $1:2$, we have: $x = \frac{1(-2t) + 2(t)}{1 + 2} = \frac{-2t + 2t}{3} = 0$ $y = \frac{1(-8t^3) + 2(t^3)}{1 + 2} = \frac{-8t^3 + 2t^3}{3} = \frac{-6t^3}{3} = -2t^3$ So, the point is $R(0, -2t^3)$.

Step 6: Identify the ordinate of the point.

• Why: The problem specifically asks for the y-coordinate.
• Calculation: The ordinate (y-coordinate) of the point is $-2t^3$.

Common Mistakes & Tips

• Factoring the Cubic Equation: Be careful when factoring the cubic equation. Remember that since the line is tangent at $P$, $(x-t)$ will be a repeated factor.
• Section Formula: Ensure you apply the section formula correctly, using the correct ratio and coordinates.
• Alternative Method using Roots of Polynomials: The equation of the tangent at $P(t, t^3)$ is $y - t^3 = 3t^2(x - t)$. Substituting $y=x^3$ into this equation gives: $x^3 - t^3 = 3t^2(x - t)$ or $x^3 - 3t^2x + 2t^3 = 0$. This is a cubic equation whose roots are the x-coordinates of the intersection points. We know that the tangent touches the curve at P, so $x=t$ is a double root. Let the roots be $t, t, x_Q$. The sum of roots is $t + t + x_Q = 0 \Rightarrow x_Q = -2t$.

Summary

We found the derivative of the curve, used it to find the equation of the tangent at point P, and then found the intersection point Q of the tangent and the curve. Finally, we used the section formula to find the ordinate of the point that divides PQ internally in the ratio 1:2, which is $-2t^3$.

Final Answer The final answer is \boxed{-2t^3}, which corresponds to option (C).