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.
What is that variable a in f(a)?
I do it like that and think it is quite fast:
y' at (3,9) = 6
slope of normal at (3,9) = -1/6
Thus required equation is x+6y = 3 + 6(9)
idk if this is fast or not actually...
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