# Equation of Tangent and Normal to the Curve

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 equations of the tangent and the normal at the point ‘t’ on the curve x = $$a sin^3 t$$, y = $$b cos^3 t$$.

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