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To find the equation of the tangent to the curve at the given point corresponding to \( t = 9 \), we need to follow these steps:

### Given Parametric Equations:
\[
x = \sqrt{t}, \quad y = t^2 - 2t
\]
We are asked to find the tangent at \( t = 9 \).

### Step 1: Find the point on the curve corresponding to \( t = 9 \).

1. **x-coordinate**:
   \[
   x(9) = \sqrt{9} = 3
   \]
   
2. **y-coordinate**:
   \[
   y(9) = 9^2 - 2(9) = 81 - 18 = 63
   \]
   
So, the point on the curve is \( (3, 63) \).

### Step 2: Find the slope of the tangent (dy/dx) using parametric derivatives.

The formula for the slope \( \frac{dy}{dx} \) in parametric form is:
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]

Now, calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):

- **\( \frac{dx}{dt} \)**:
  \[
  \frac{dx}{dt} = \frac{d}{dt} (\sqrt{t}) = \frac{1}{2\sqrt{t}}
  \]

- **\( \frac{dy}{dt} \)**:
  \[
  \frac{dy}{dt} = \frac{d}{dt}(t^2 - 2t) = 2t - 2
  \]

### Step 3: Evaluate \( \frac{dy}{dx} \) at \( t = 9 \).

1. **\( \frac{dx}{dt} \) at \( t = 9 \)**:
   \[
   \frac{dx}{dt} = \frac{1}{2\sqrt{9}} = \frac{1}{6}
   \]

2. **\( \frac{dy}{dt} \) at \( t = 9 \)**:
   \[
   \frac{dy}{dt} = 2(9) - 2 = 18 - 2 = 16
   \]

Now, compute \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{16}{\frac{1}{6}} = 16 \times 6 = 96
\]

### Step 4: Equation of the Tangent Line.

The slope of the tangent line at \( t = 9 \) is 96, and the point is \( (3, 63) \). Using the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
where \( m = 96 \), \( (x_1, y_1) = (3, 63) \), the equation becomes:
\[
y - 63 = 96(x - 3)
\]
Simplifying:
\[
y - 63 = 96x - 288
\]
\[
y = 96x - 225
\]

Thus, the equation of the tangent is:
\[
y = 96x - 225
\]

---

Would you like more details or explanations on any step?

Here are some related questions to consider:
1. How is the formula for the slope of a parametric curve derived?
2. What is the geometric interpretation of the tangent line in this context?
3. Could we have found the tangent line without parametric differentiation?
4. How does the curve behave for other values of \( t \)?
5. How would we find the normal line instead of the tangent?

**Tip:** When working with parametric equations, always remember to find both \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) to compute the correct slope \( \frac{dy}{dx} \).

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter: x = √t, y = t^2 - 2t; t = 9.

Learn how to find the equation of the tangent to the parametric curve defined by x = √t and y = t^2 - 2t at t = 9. This problem covers parametric differentiation and tangent line equations. Ideal for Grades 11-12 math students.

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