Key Concepts and Formulas

• Parametric Representation of a Curve: A curve defined by equations $x = f(t)$ and $y = g(t)$, where $t$ is a parameter.
• Slope of the Tangent to a Parametric Curve: The derivative $\frac{dy}{dx}$ for a parametric curve is given by the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$
• Equation of a Tangent Line: The equation of a tangent line at point $(x_0, y_0)$ with slope $m$ is $y - y_0 = m(x - x_0)$.
• Slope of a Line (Chord): Given two points $(x_1, y_1)$ and $(x_2, y_2)$, the slope of the line passing through them is $\frac{y_2 - y_1}{x_2 - x_1}$.

Step-by-Step Solution

Step 1: Understand the Curve and Identify Point P

The curve is given by the parametric equations: $x = 4t^2 + 3$ $y = 8t^3 - 1$ For a specific value of the parameter $t$, the point P on the curve is $P(t) = (4t^2 + 3, 8t^3 - 1)$. This is our initial point of tangency.

Step 2: Calculate the Slope of the Tangent at Point P

To find the slope of the tangent line at P, we need to calculate $\frac{dy}{dx}$.

• Step 2a: Differentiate $x$ with respect to $t$ $\frac{dx}{dt} = \frac{d}{dt}(4t^2 + 3) = 8t$ Explanation: This gives us the instantaneous rate of change of the x-coordinate with respect to the parameter $t$.

• Step 2b: Differentiate $y$ with respect to $t$ $\frac{dy}{dt} = \frac{d}{dt}(8t^3 - 1) = 24t^2$ Explanation: This gives us the instantaneous rate of change of the y-coordinate with respect to the parameter $t$.

• Step 2c: Calculate $\frac{dy}{dx}$ using the Chain Rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{24t^2}{8t} = 3t$ Explanation: This is the slope of the tangent line to the curve at the point P, corresponding to the parameter $t$.

Step 3: Represent Point Q and Calculate the Slope of Chord PQ

Let Q be another point on the curve, corresponding to a different parameter, say $\lambda$. Explanation: We use a new parameter $\lambda$ to denote point Q because Q is distinct from P (unless $\lambda$ happens to be equal to $t$). This helps us differentiate between the two points.

The coordinates of Q are $Q(\lambda) = (4\lambda^2 + 3, 8\lambda^3 - 1)$.

Now, we calculate the slope of the chord connecting P and Q: $\text{Slope of PQ} = \frac{y_Q - y_P}{x_Q - x_P}$ $= \frac{(8\lambda^3 - 1) - (8t^3 - 1)}{(4\lambda^2 + 3) - (4t^2 + 3)}$ $= \frac{8\lambda^3 - 8t^3}{4\lambda^2 - 4t^2}$ Factor out common terms and use the difference of cubes ($a^3 - b^3 = (a-b)(a^2+ab+b^2)$) and difference of squares ($a^2 - b^2 = (a-b)(a+b)$) identities: $= \frac{8(\lambda^3 - t^3)}{4(\lambda^2 - t^2)}$ $= \frac{2(\lambda - t)(\lambda^2 + \lambda t + t^2)}{(\lambda - t)(\lambda + t)}$ Explanation: Factoring is a critical step here. It allows us to simplify the expression and will later help us identify the parameter for Q.

Step 4: Equate Slopes and Solve for $\lambda$

Since the tangent at P meets the curve again at Q, the slope of the chord PQ must be equal to the slope of the tangent at P.

$\text{Slope of PQ} = \text{Slope of tangent at P}$ $\frac{2(\lambda - t)(\lambda^2 + \lambda t + t^2)}{(\lambda - t)(\lambda + t)} = 3t$

Since Q is a point other than P, we know that $\lambda \neq t$. Therefore, we can safely cancel the term $(\lambda - t)$ from both the numerator and the denominator: $\frac{2(\lambda^2 + \lambda t + t^2)}{\lambda + t} = 3t$ Explanation: Cancelling $(\lambda - t)$ is valid because if $\lambda = t$, Q would be the same point as P, which contradicts the problem statement that the tangent meets the curve again at Q.

Now, we solve for $\lambda$: $2(\lambda^2 + \lambda t + t^2) = 3t(\lambda + t)$ $2\lambda^2 + 2\lambda t + 2t^2 = 3\lambda t + 3t^2$ Rearrange the terms to form a quadratic equation in $\lambda$: $2\lambda^2 + (2t - 3t)\lambda + (2t^2 - 3t^2) = 0$ $2\lambda^2 - t\lambda - t^2 = 0$ Explanation: We rearrange the equation to have all terms on one side equal to zero.

Now, we solve this quadratic equation for $\lambda$. We can factor the quadratic: $2\lambda^2 - 2t\lambda + t\lambda - t^2 = 0$ $2\lambda(\lambda - t) + t(\lambda - t) = 0$ $(2\lambda + t)(\lambda - t) = 0$ The solutions are $\lambda = t$ and $\lambda = -\frac{t}{2}$. Since $\lambda \neq t$, we take $\lambda = -\frac{t}{2}$. Explanation: We solve the quadratic equation to find the value of $\lambda$ for point Q. We discard $\lambda=t$ as Q and P must be distinct.

Step 5: Find the Coordinates of Point Q

Now that we have the value of $\lambda$, we can find the coordinates of Q by substituting $\lambda = -\frac{t}{2}$ into the parametric equations: $x_Q = 4\lambda^2 + 3 = 4\left(-\frac{t}{2}\right)^2 + 3 = 4\left(\frac{t^2}{4}\right) + 3 = t^2 + 3$ $y_Q = 8\lambda^3 - 1 = 8\left(-\frac{t}{2}\right)^3 - 1 = 8\left(-\frac{t^3}{8}\right) - 1 = -t^3 - 1$ Therefore, the coordinates of Q are $(t^2 + 3, -t^3 - 1)$.

Common Mistakes & Tips

• Don't forget the chain rule: When finding $\frac{dy}{dx}$ for parametric equations, remember to use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
• Simplify carefully: Be meticulous when simplifying algebraic expressions, especially when factoring.
• Remember the condition $\lambda \neq t$: The points P and Q are distinct, so their parameters must be different. This allows you to cancel $(\lambda - t)$.

Summary

We found the coordinates of point Q by first finding the slope of the tangent at point P using the chain rule. Then, we found the slope of the chord PQ and equated it to the slope of the tangent at P. Solving for $\lambda$, the parameter of point Q, and substituting it into the parametric equations gave us the coordinates of Q as $(t^2 + 3, -t^3 - 1)$.

Final Answer

The final answer is $\boxed{(t^2 + 3, -t^3 - 1)}$, which corresponds to option (A).