# Tangents and Normals

## Equation of Tangents of a Function

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 $y' = 2$ => $x = \frac{1}{16}$

## Related Topics

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