Here you will learn equation of tangent and normal to the curve with examples.

Let’s begin –

## Equation of Tangent and Normal to the Curve

We know that the equation of line passing through a point **\((x_1, y_1)\)** and having slope m is **\(y – y_1\) = m\((x – x_1)\).**

and we know that the slopes of the tangent and the normal to the curve y = f(x) at a point P\((x_1, y_1)\) are **\(({dy\over dx})_P\)** and **-\(1\over ({dy\over dx})_P\)** respectively.

Therefore the **equation of the tangent **at P\((x_1, y_1)\) to the curve y = f(x) is

\(y – y_1\) = \(({dy\over dx})_P\) (\(x – x_1\))

Since the normal at P\((x_1, y_1)\) passes through P and has slope -\(1\over ({dy\over dx})_P\).

Therefore, the **equation of the normal** at P\((x_1, y_1)\) to the curve y = f(x) is

\(y – y_1\) = \(-1\over ({dy\over dx})_P\) (\(x – x_1\))

**Note :**

1). If \(({dy\over dx})_P\) = \(\pm \infty\), then the tangent at \((x_1, y_1)\) is parallel to y-axis and its equation is x = \(x_1\).

2). If \(({dy\over dx})_P\) = 0, then the normal at \((x_1, y_1)\) is parallel to y-axis and its equation is x = \(x_1\).

3). If \(({dy\over dx})_P\) = \(\pm \infty\), then the normal at \((x_1, y_1)\) is parallel to x-axis and its equation is y = \(y_1\).

4). If \(({dy\over dx})_P\) = 0, then the tangent at \((x_1, y_1)\) is parallel to x-axis and its equation is y = \(y_1\).

**Example** : find the equation of the tangent to curve y = \(-5x^2 + 6x + 7\) at the point (1/2, 35/4).

**Solution** : The equation of the given curve is

y = \(-5x^2 + 6x + 7\)

\(\implies\) \(dy\over dx\) = -10x + 6

\(\implies\) \(({dy\over dx})_{(1/2, 35/4)}\) = \(-10\over 4\) + 6 = 1

The required equation at (1/2, 35/4) is

y – \(35\over 4\) = \(({dy\over dx})_{(1/2, 35/4)}\) \((x – {1\over 2})\)

\(\implies\) y – 35/4 = 1(x – 1/2)

\(\implies\) y = x + 33/4

### Related Questions

Find the equation of the normal to the curve y = \(2x^2 + 3 sin x\) at x = 0.