To solve this problem, let's go through each step to find the equations of the tangents.

Given Information

The parametric equations of the curve are: $x = 9t^2 + 3$ $y = 6t^3 + 3$ We are asked to find the tangent lines that pass through the point $(12, 9)$.

Step 1: Find $\frac{dy}{dx}$ in Terms of $t$

To get the slope of the tangent line, we need to compute $\frac{dy}{dx}$, which can be done using: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

1. Differentiate $x$ with respect to $t$: $\frac{dx}{dt} = 18t$

2. Differentiate $y$ with respect to $t$: $\frac{dy}{dt} = 18t^2$

So, $\frac{dy}{dx} = \frac{18t^2}{18t} = t$ Thus, the slope of the tangent line at a point on the curve is given by $t$.

Step 2: Find the Points on the Curve

To determine the points on the curve that might give us tangents passing through $(12, 9)$, we equate $x$ and $y$ expressions to match this point.

1. Solve for $t$ when $x = 12$: $9t^2 + 3 = 12$ $9t^2 = 9$ $t^2 = 1$ $t = \pm 1$

2. Evaluate $y$ at $t = 1$ and $t = -1$:

   • For $t = 1$: $y = 6(1)^3 + 3 = 6 + 3 = 9$ So, the point $(x, y) = (12, 9)$ is on the curve when $t = 1$.

   • For $t = -1$: $y = 6(-1)^3 + 3 = -6 + 3 = -3$ So, for $t = -1$, the point on the curve is $(x, y) = (12, -3)$.

Step 3: Find the Tangent Lines

Now we have two points on the curve where a tangent might pass through $(12, 9)$:

1. At $t = 1$:

   • Slope of the tangent $m = t = 1$.
   • Equation of the tangent line in point-slope form: $y - 9 = 1(x - 12)$ Simplifying: $y = x - 3$

2. At $t = -1$:

   • Slope of the tangent $m = t = -1$.
   • Equation of the tangent line: $y + 3 = -1(x - 12)$ Simplifying: $y = -x + 9$

Final Answer

The equations of the tangent lines that pass through $(12, 9)$ are: $y = x - 3, \quad y = -x + 9$

Would you like further explanation or have any questions?

Related Questions

1. How can we find the slope of the tangent line at any point on a parametric curve?
2. What are the steps to determine whether a line passes through a given point?
3. How do we derive the equation of a tangent line to a parametric curve?
4. How can we verify that the point $(12, 9)$ lies on both tangent lines?
5. What role does parametric differentiation play in finding slopes of curves?

Tip

When working with parametric equations, always remember to differentiate $x$ and $y$ with respect to $t$ before finding $\frac{dy}{dx}$. This makes it easier to determine the slope of the tangent line at a given point.