Find the equation of the normal line, $$ y=m_{n} x+C $$ at the same point.

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Suppose we have a curve given by a differentiable function $$ y=f(x) $$ and a point on the curve $$(a, f(a))$$. The tangent line at that point has slope

$$
m_{t} = f'(a).
$$

Because the normal line is perpendicular to the tangent line, its slope $$ m_{n} $$ is the negative reciprocal of $$ m_{t} $$:

$$
m_{n} = -\frac{1}{f'(a)} \quad \text{(provided } f'(a) \neq 0\text{)}.
$$

To write the equation of the normal line in point-slope form, we use the point $$(a, f(a))$$:

$$
y - f(a) = -\frac{1}{f'(a)} \left( x - a \right).
$$

If you prefer the slope-intercept form $$ y = m_{n}x + C $$, we solve for $$ C $$ by rearranging the equation:

$$
y = -\frac{1}{f'(a)} \, x + \left[ f(a) + \frac{a}{f'(a)} \right].
$$

Thus, the equation of the normal line at $$(a, f(a))$$ is

$$
y = -\frac{1}{f'(a)} \, x + \left( f(a) + \frac{a}{f'(a)} \right).
$$
The equation of the normal line at point $$(a, f(a))$$ is $$ y = -\frac{1}{f'(a)}x + \left( f(a) + \frac{a}{f'(a)} \right) $$.

Find the equation of the normal line, at the same point.

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