How to do this quickly?

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

Find the equation of the normal to y= x^{2} 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.

Guest Aug 10, 2017

#1**+1 **

What is that variable a in f(a)?

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

y=x^{2}

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...

Guest Aug 10, 2017

edited by
Guest
Aug 10, 2017

#2**+1 **

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

CPhill
Aug 10, 2017