What is the normal of y=sqrt(x),(9,3) ?

The normal of y=sqrt(x),(9,3) is y=-6x+57

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normal of y=sqrt(x),(9,3)

normal of

y = x, (9, 3)

y = - 6 x + 57

Solution steps

Find the slope of

y = x : \frac{d y}{d x} = \frac{1}{2 x}

Compute the slope of

y = x

(9, 3) : m = \frac{1}{6}

Compute the slope of the perpendicular line:

m_{p} = - 6

Find the line with slope m=

- 6

and passing through

(9, 3) : y = - 6 x + 57

y = - 6 x + 57

\frac{sin ( x ) + cos ( x )}{cos ( x )}

y = \frac{8 x}{x ^{2} + 1}, (- 1, - 4)

\int \frac{1}{x ^{2} + 2 x - 24} d x

\frac{\partial}{\partial y} (\frac{( x ^{2} y )}{x ^{2} + y ^{2}})

6 x^{2} + 2 x y + 2 y^{2} = 74, (3, 2)

What is the normal of y=sqrt(x),(9,3) ?
The normal of y=sqrt(x),(9,3) is y=-6x+57