https://file.aiquickdraw.com/mathbot/file/1730777781172z8szqx7t.jpg

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.

Find equations of the tangents to the curve x = 9t^2 + 3, y = 6t^3 + 3 that pass through the point (12, 9).

Discover how to find the equations of tangent lines to the parametric curve x = 9t^2 + 3, y = 6t^3 + 3 that pass through the point (12, 9). This problem covers parametric differentiation, calculating slopes, and forming tangent equations. Suitable for advanced high school math students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation