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 y–y1 = m(x–x1).
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
y–y1 = (dydx)P (x–x1)
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
y–y1 = −1(dydx)P (x–x1)
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) (x–12)
⟹ y – 35/4 = 1(x – 1/2)
⟹ y = x + 33/4
Related Questions
Find the equation of the normal to the curve y = 2x2+3sinx at x = 0.