To determine the equation of the tangent line to the curve defined parametrically by:
$x = t^2 - 4t \quad \text{and} \quad y = t^2 + 4t + 1$
at the given point $(0, 33)$, we proceed as follows:

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Step 1: Find the parameter $t$ at the given point $(0, 33)$

The $x$-coordinate is given as $0$, so we solve for $t$ using:
$x = t^2 - 4t = 0$
Factorizing:
$t(t - 4) = 0 \implies t = 0 \, \text{or} \, t = 4$
Now check which value of $t$ corresponds to $y = 33$:

• For $t = 0$:
  $y = t^2 + 4t + 1 = 0^2 + 4(0) + 1 = 1 \quad (\text{not 33}).$

• For $t = 4$:
  $y = t^2 + 4t + 1 = 4^2 + 4(4) + 1 = 16 + 16 + 1 = 33.$
  Thus, $t = 4$ corresponds to the point $(0, 33)$.

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Step 2: Differentiate $x$ and $y$ with respect to $t$

To find the slope of the tangent line, we compute $dy/dx$ using the chain rule:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
We first compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

• From $x = t^2 - 4t$:
  $\frac{dx}{dt} = 2t - 4$

• From $y = t^2 + 4t + 1$:
  $\frac{dy}{dt} = 2t + 4$

Thus, the slope of the tangent line is:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + 4}{2t - 4}.$

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Step 3: Compute the slope at $t = 4$

Substitute $t = 4$ into $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{2(4) + 4}{2(4) - 4} = \frac{8 + 4}{8 - 4} = \frac{12}{4} = 3.$
Thus, the slope of the tangent line at $t = 4$ is $3$.

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Step 4: Find the equation of the tangent line

The tangent line passes through the point $(0, 33)$ and has a slope of $3$. Using the point-slope form of a line:
$y - y_1 = m(x - x_1),$
where $m = 3$, $x_1 = 0$, and $y_1 = 33$:
$y - 33 = 3(x - 0).$
Simplify:
$y = 3x + 33.$

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Final Answer:

The equation of the tangent line is:
$y = 3x + 33.$

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Do you want more details, or have any other questions?

Related Questions:

1. How do you find the slope of a tangent line for parametric curves?
2. What is the chain rule and how is it applied to parametric equations?
3. How do you check which $t$-value corresponds to a given point on a curve?
4. Can you find the normal line to the curve at the same point?
5. How would you sketch the curve and the tangent line on a graph?

Tip:

Always verify which parameter value ($t$) corresponds to the given point before proceeding with derivatives.