Finding Tangent line from Parametric

Question

I need to find an equation of the tangent line to the curve $x=5+t^2-t$, $y=t^2+5$ at the point $(5,6)$.

Setting $x=5$ and $y = 6$ and solving for $t$ gives me $t=0,1,-1$. I know I have to do y/x, and then take the derivative. But how do I know what $t$ value to use?

Answer 1

You have

i) $x=5+t^2-t$ and ii) $y=t^2+5$ and $P=(5,6)$.

From $P=(5,6)$ you get $x=5$ and $y=6$.

From ii) you get now $t^2=1\Rightarrow t=1$ or $t=-1$ and from i) you get (knowing that $t\in${1,-1} ):$\ $ $t=1$

Your curve has the parametric representation

$\gamma: I\subseteq\mathbb{R}\rightarrow {\mathbb{R}}^2: t\mapsto (5+t^2-t,t^2+5)$.

Therefore, $\frac{d}{dt}\gamma=(2t-1,2t)$.

Put $t=1$ into the derivative of $\gamma$ and from this you can get the slope of the tangent in P.

Answer 2

Please type up or paste the image text in your problem statement.

Solving for $t: x(t)=5$ gives $t \in \{0,1\}$ and solving for $t:y(t) = 6$ gives $t \in \{ 1,-1\}$. The common $t$ is $t=1$. Compute $dy/dx$ and not $y/x$ at $t=1$ , and substitute this into the slope of the tangent.