Find the tangent line at pont $q=(a,b)$ of the following curves.

Question

I need help with this problem:

Find, where appropriate the equation of the tangent line to $C$ at the point $q=(a,b)$ on $C$. Indicate points on $C$ where no tangent line exists.

1. the $y$-axis, $x=0$.
2. the parabola $y+1=(x-2)^2$.
3. the circle $(x-1)^2+(y-2)^2=4$

For 1. I used the parametrization $f(t)=(0,t)$ and the derivative at $a$ is $f'(a)=(0,1)$. For 2. I used the parametrization $f(t)=(t+2,t^2-1)$ and $f'(a)=(1, 2t)$ and for 3. I used $f(t)=(2\cos t+1, 2\sin t+2)$ and $f'(a)=(-2\sin t, 2\cos t)$.

How do I find the equation of the tangent line at q?

Answer 1

The equation of the tangent line is $y-y_0=y'(x_0)(x-x_0)$ or $x-x_0=x'(y_0)(y-y_0)$.

1

$$x=0\implies x'=0.$$ Thus we have that the tangent line at $(0,y_0)$ is $x=0$.

2

$$y=(x-2)^2 -1\implies y'=2x-4.$$ Thus, the tangent line at $(x_0,y_0)$ is $$y-y_0=(2x_0-4)(x-x_0)$$ where $y_0=(x_0-2)^2 -1.$

3

$$(x-1)^2 +(y-2)^2=1 \implies x-1+(y-2)y'=0, (x-1)x'+y-2=0.$$

So, if $y\ne 2$ then $y'=-\frac{x-1}{y-2}$ and if $x\ne 1$ then $x'=-frac{y-2}{x-1}.$