What is the normal of y=x^2,(3,9) ?

The normal of y=x^2,(3,9) is y=-1/6 x+19/2

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normal of y=x^2,(3,9)

normal of

y = x^{2}, (3, 9)

y = - \frac{1}{6} x + \frac{19}{2}

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 6

Compute the slope of the perpendicular line:

m_{p} = - \frac{1}{6}

Find the line with slope m=

- \frac{1}{6}

and passing through

(3, 9) : y = - \frac{1}{6} x + \frac{19}{2}

y = - \frac{1}{6} x + \frac{19}{2}