# Step By Step Calculus » 13.2 - Tangent Lines for Curves

Synopsis
The tangent line to a curve y=f(x)y=f(x) at x=ax=a is same as the linear approximation L_a(x)L_a(x) of y=f(x)y=f(x) at x=a,x=a, i.e. the tangent lines are:
\displaystyle y=f(a)+f'(a)(x-a).
\displaystyle y=f(a)+f'(a)(x-a).
For finding tangent lines to a curve defined either explicitly as y=f(x),y=f(x), or implicitly as g(x,y)=a,g(x,y)=a, or parametrically as x=x(t),y=y(t)x=x(t),y=y(t) we can use the following steps.
Step 1: (Differentiate) Find the first derivative dy/dx.dy/dx.
Step 2: (Evaluate) Evaluate the xx at which one wants to find the tangent lines, say x_1.x_1. There can be more than one value of x.x. Consider each x_1x_1 separately.
Then, evaluate the function yy at x_1x_1 and say it is evaluated to y_1.y_1. There can be more than one y_1y_1 for a x_1.x_1. Consider each pair (x_1,y_1)(x_1,y_1) separately.
Finally, evaluate the first derivative at (x_1, y_1).(x_1, y_1).
Step 3: (Substitute) Plugging \displaystyle x_1, y_1, \left.\frac{dy}{dx}\right|_{x=x_1, y=y_1}\displaystyle x_1, y_1, \left.\frac{dy}{dx}\right|_{x=x_1, y=y_1} in the formula, we get the equation of the tangent line as
\displaystyle y=y_1+(x-x_1)\left.\frac{dy}{dx}\right|_{x=x_1, y=y_1}.
\displaystyle y=y_1+(x-x_1)\left.\frac{dy}{dx}\right|_{x=x_1, y=y_1}.
For a parametrically defined curvex=x(t), y=y(t),x=x(t), y=y(t), the first derivative or slope of the tangent is defined by
\displaystyle \frac{dy}{dx}=\frac{y'(t)}{x'(t)}.
\displaystyle \frac{dy}{dx}=\frac{y'(t)}{x'(t)}.