To find the equation of the tangent line to the given parametric curve at \( t = 1 \), we need to determine the slope of the tangent line and the point of tangency. The slope of the tangent line can be found by differentiating \( y \) with respect to \( x \) using the chain rule. The point of tangency is obtained by substituting \( t = 1 \) into the parametric equations.
Given the parametric equations:
\[
x = 2t + 4
\]
\[
y = 8t^2 - 2t + 4
\]
To find the slope of the tangent line, we need to find \(\frac{dy}{dx}\). Using the chain rule for parametric equations, we have:
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]
First, calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):
\[
\frac{dx}{dt} = \frac{d}{dt}(2t + 4) = 2
\]
\[
\frac{dy}{dt} = \frac{d}{dt}(8t^2 - 2t + 4) = 16t - 2
\]
Now, substitute these into the expression for \(\frac{dy}{dx}\):
\[
\frac{dy}{dx} = \frac{16t - 2}{2} = 8t - 1
\]
Substitute \( t = 1 \) into \(\frac{dy}{dx}\) to find the slope of the tangent line at this point:
\[
\frac{dy}{dx}\bigg|_{t=1} = 8(1) - 1 = 7
\]
Substitute \( t = 1 \) into the parametric equations to find the point of tangency:
\[
x = 2(1) + 4 = 6
\]
\[
y = 8(1)^2 - 2(1) + 4 = 10
\]
The point on the curve at \( t = 1 \) is \((6, 10)\).
Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point of tangency:
\[
y - 10 = 7(x - 6)
\]
Simplify to get the equation of the tangent line:
\[
y - 10 = 7x - 42
\]
\[
y = 7x - 32
\]
From the parametric equation for \( x \):
\[
x = 2t + 4 \implies t = \frac{x - 4}{2}
\]
Substitute \( t = \frac{x - 4}{2} \) into the equation for \( y \):
\[
y = 8\left(\frac{x - 4}{2}\right)^2 - 2\left(\frac{x - 4}{2}\right) + 4
\]
Simplify:
\[
y = 8\left(\frac{(x - 4)^2}{4}\right) - \frac{2(x - 4)}{2} + 4
\]
\[
y = 2(x - 4)^2 - (x - 4) + 4
\]
\[
y = 2(x^2 - 8x + 16) - x + 4 + 4
\]
\[
y = 2x^2 - 16x + 32 - x + 8
\]
\[
y = 2x^2 - 17x + 40
\]
Differentiate the Cartesian equation:
\[
\frac{dy}{dx} = \frac{d}{dx}(2x^2 - 17x + 40) = 4x - 17
\]
Evaluate at \( x = 6 \):
\[
\frac{dy}{dx}\bigg|_{x=6} = 4(6) - 17 = 24 - 17 = 7
\]
The slope is consistent with the parametric form. The point \((6, 10)\) is on the curve, so the tangent line is:
\[
y - 10 = 7(x - 6)
\]
\[
y = 7x - 32
\]
a. The equation of the tangent line to the curve at \( t = 1 \) without eliminating the parameter is:
\[
\boxed{y = 7x - 32}
\]
b. The equation of the tangent line by eliminating the parameter is:
\[
\boxed{y = 7x - 32}
\]
Answered
