Answer
$y=24x-40$
Work Step by Step
$\qquad\qquad\qquad\qquad\qquad\qquad$Find the point at which $t = 4$
$x|_{t=4}=\sqrt{4}=2$
$y|_{t=4}=(4)^2-2(4)=16-8=8$
Point: $(2,8)$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$Now find $\displaystyle \frac{dy}{dx}$
$\displaystyle \frac{dy}{dt}=2t-2$
$\displaystyle \frac{dx}{dt}=\frac{1}{2\sqrt{t}}$
$\displaystyle \frac{dy}{dx}=\frac{2t-2}{\frac{1}{2\sqrt{t}}}=2\sqrt{t}(2t-2)=4\sqrt{t}(t-1)$
$\qquad\qquad\qquad\qquad$Now plug $t=4$ in for $t$ and find the slope of the line:
$\displaystyle \frac{dy}{dx}|_{t=4}=4\sqrt{4}(4-1)=4(2)(3)=24$
$\qquad\qquad\qquad\qquad$Now use the point-slope form of a line to
$\qquad\qquad\qquad\qquad$find the equation tangent to the parametric curve.
$y-y_1=m(x_1-x)\quad$
$y-8=24(x-2)$
$y=24x-48+8$
$y=24x-40$
