We know from coordinate geometry that the the equation of a straight line passing through a given point $(x_0, y_0)$ having finite slope m is given by
$y – y_0= m (x – x_0)$
Now we also know from Derivatives of the function that $\frac {dy}{dx}_{x=x_0}$ represent the slope of the tangent of the function at point $(x_0, y_0)$
Therefore Equation of Tangent of a Function
$y – y_0= f'(x_0) (x – x_0)$

Equation of Normals of a Function

We know that The normal to the curve at a point is perpendicular to the tangent at that point
So slope of the Normal would be $\frac {-1}{f'(x_0)}$
Hence equation of Normal of a Function
$y – y_0= \frac {-1}{f'(x_0)} (x – x_0)$
$(y – y_0)f'(x_0) +(x – x_0)=0$

Solved Examples

Question 1
Find the equation of the tangent to the curve \( y = x^2 - 4x + 3 \) at the point where \( x = 2 \). Solution
\( y = x^2 - 4x + 3 \)
\( y' = 2x - 4 \)
For \( x = 2 \), \( y' = 0 \) and \( y = -1 \)
Therefore Equation of tangent
\( y + 1 = 0(x - 2) \)
or
y = -1

Question 2
Find the equation of the normal to the curve \( y = 3x^2 - 5x + 2 \) at the point where \( x = 1 \). Solution
\( y = 3x^2 - 5x + 2 \)
\( y' = 6x - 5 \)
For \( x = 1 \), \( y' = 1 \) and \( y = 0 \)
Therefore Equation of normal
\( y = -x + 1 \)

Question 3
At what points on the curve \( y = 2x^2 - 4x - 5 \) is the slope of the tangent equal to 4? Solution
\( y = 2x^2 - 4x - 5 \)
\( y' = 4x - 4 \)
Equating \( y' = 4 \) => \( x = 2 \).
At this point, the slope of the tangent is 4.

Question 4
Find the equation of the normal to the curve \( y = \sqrt{x} \) which has a slope of -1/2. Solution
\( y = \sqrt{x} \)
\( y' = \frac{1}{2\sqrt{x}} \)
For the slope of normal = \( -\frac{1}{2} \), the slope of tangent = 2.
Equating \( y' = 2 \) => \( x = \frac{1}{16} \)