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Find an equation of the normal line to the curve y = x that is parallel to the line 2x y = 1.

Sagot :

The equation of the normal line is y = - 2x + 3

We know that,

The equation of normal line given of the curve is given by

( y - [tex]y_{1}[/tex] ) = m ( x - [tex]x_{1}[/tex] )

Given that,

Equation of curve is y = [tex]\sqrt{x}[/tex]

Equation of the line parallel to the normal line of curve is 2x + y = 1

The slope of normal is same as that of the line parallel to it. So,

2x + y = 1

y = - 2x + 1

On differentiating both sides by x,

dy / dx = - 2

which is the slope of normal line

To find slope of the curve,

y = [tex]\sqrt{x}[/tex]

y = [tex]x^{1/2}[/tex]

dy / dx = 1/2 x

which is the slope of the curve.

Since the curve and its normal line is perpendicular,

[tex]m_{1} m_{2}[/tex] = -1

-2 * 1/2 x = -1

x = 1

At x = 1,

y = [tex]\sqrt{1}[/tex] = 1

The point is ( 1 , 1 )

The equation of normal line given of the curve is

( y - [tex]y_{1}[/tex] ) = m ( x - [tex]x_{1}[/tex] )

y - 1 = - 2 ( x - 1 )

y = - 2x + 3

Therefore, The equation of the normal line is y = - 2x + 3

To know more about equation of the normal line

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