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# How to do this quickly, find equation of normal.

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How to do this quickly?

The slow way: f(a) - (x-a)/f'(a)

Find the equation of the normal to y= x2 at the point (3, 9)

What is the fastest way to do this? The one where the answer you get would be x + 6y = 57

The slow way answer is y = 9.5 - x/6

Same answer but I belive the fast one is, well, faster. But I forgot how to do it.

Aug 10, 2017

#1
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What is that variable a in f(a)?

I do it like that and think it is quite fast:

y=x2

y'= 2x

y' at (3,9) = 6

slope of normal at (3,9) = -1/6

Thus required equation is x+6y = 3 + 6(9)

x+6y=57

idk if this is fast or not actually...

Aug 10, 2017
edited by Guest  Aug 10, 2017
#2
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For the function x^2, the slope of the tangent line to the curve at any point on the curve is 2x

A line perpendicuar to this tangent line will have the slope of  - 1 / [2x ]

So....at (3,9)....the slope of a perpendicular line will be    -1 / [ 2 (3) ]  =  -1/6

So...the equation of a normal line at (3,9) will be

y  = -(1/6) ( x - 3) + 9    =   (-1/6)x + 1/2 + 9    =   (-1/6)x + 19/2

In standard form we would have

-6y  = x + 57

x - 6y + 57  = 0

Here's the graph : https://www.desmos.com/calculator/m8bovfpy9w

Aug 10, 2017