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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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