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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 (x1,y1) and having slope m is yy1 = m(xx1).

and we know that the slopes of the tangent and the normal to the curve y = f(x) at a point P(x1,y1) are (dydx)P and -1(dydx)P respectively. 

Therefore the equation of the tangent at P(x1,y1) to the curve y = f(x) is

yy1 = (dydx)P (xx1)

Since the normal at P(x1,y1) passes through P and has slope -1(dydx)P.

Therefore, the equation of the normal at P(x1,y1) to the curve y = f(x) is

yy1 = 1(dydx)P (xx1)

Note :

1). If (dydx)P = ±, then the tangent at (x1,y1)  is parallel to y-axis and its equation is x = x1.

2). If (dydx)P = 0, then the normal at (x1,y1)  is parallel to y-axis and its equation is x = x1.

3). If (dydx)P = ±, then the normal at (x1,y1)  is parallel to x-axis and its equation is y = y1.

4). If (dydx)P = 0, then the tangent at (x1,y1)  is parallel to x-axis and its equation is y = y1.

Example : find the equation of the tangent to curve y = 5x2+6x+7  at the point (1/2, 35/4).

Solution : The equation of the given curve is 

y = 5x2+6x+7 

dydx = -10x + 6

(dydx)(1/2,35/4) = 104 + 6 = 1

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

y – 354 = (dydx)(1/2,35/4) (x12)

y – 35/4 = 1(x – 1/2)

y = x + 33/4


Related Questions

Find the equations of the tangent and the normal at the point ‘t’ on the curve x = asin3t, y = bcos3t.

Find the equation of the normal to the curve y = 2x2+3sinx at x = 0.

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