Answer
(a) $y=-\frac{1}{4}x+\frac{3}{4}$
(b) $y=-\frac{1}{4}x+\frac{3}{4}$
Work Step by Step
Part (a)
Find the slope of the tangent to the curve:
$m=\frac{dy}{dx}$
$m=\frac{dy/dt}{dx/dt}$
$m=\frac{-\frac{1}{(t+4)^2}}{\frac{1}{2\sqrt{t+4}}}$
$m=-\frac{2\sqrt{t+4}}{(t+4)^2}$
$m=-\frac{2x}{x^4}|_{x=2}$ (Since $x=\sqrt{t+4}\Rightarrow x^4=(x+4)^2$)
$m=-\frac{2\cdot 2}{2^4}$
$m=-\frac{1}{4}$
Find the equation of the tangent line to the curve at $(2,1/4)$:
$y-\frac{1}{4}=m(x-2)$
$y-\frac{1}{4}=-\frac{1}{4}(x-2)$
$y-\frac{1}{4}=-\frac{1}{4}x+\frac{2}{4}$
$y=-\frac{1}{4}x+\frac{3}{4}$
Part (b)
Eliminate the parameter:
$x=\sqrt{t+4}$ and $y=\frac{1}{t+4}$
$x^2=t+4$ and $y=\frac{1}{t+4}$
$y=\frac{1}{x^2}$
Find the slope of the tangent to the curve:
$m=\frac{dy}{dx}$
$m=\frac{d}{dx}(\frac{1}{x^2})$
$m=-\frac{2}{x^3}|_{x=2}$
$m=-\frac{2}{2^3}$
$m=-\frac{1}{4}$
Find the equation of the tangent line to the curve at $(2,1/4)$:
$y-\frac{1}{4}=m(x-2)$
$y-\frac{1}{4}=-\frac{1}{4}(x-2)$
$y-\frac{1}{4}=-\frac{1}{4}x+\frac{2}{4}$
$y=-\frac{1}{4}x+\frac{3}{4}$
