normal line question

[mathmann]

1. The problem statement, all variables and given/known data
Find the equation of the normal line to the curve y(sqrd) - x(sqrd)y + 3 = 0 at the point (-2,3). (standard form)



2. Relevant equations
y - y = m(x - x)


3. The attempt at a solution

dy
__ = -6
dx

y - 3 = -6(x + 2)
6x + y +9 = 0

[neutrino]

3. The attempt at a solution

dy
__ = -6
dx

y - 3 = -6(x + 2)
6x + y +9 = 0
What you have found is the equation of the tangent line at (-2,3). You need to use the same method, but use the slope of the normal to the curve at that point. Do you know how the slopes of two mutually perpendicular lines are related to each other?

[mathmann]

Is it the reciprocal?

1
_ ?

6

[Tom Mattson]

Close, but not quite. You should look it up.

[mathmann]

its not the reciprocal?

my current answer in standard form is..

x + 6y - 16 = 0

[daniel_i_l]

What about the sign? (though it seems as though you've taken that into consideration with 1/6)

[mathmann]

I did, should be -1/6 typo above. sorry

Is it correct now?

[neutrino]

The slope of the normal is 1/6. The equation of the normal is x-6y+20=0

[mathmann]

thanks for the help. Just wanted to verify one thing though, dont you muliply -1 by the (x + 2) making them both negative.

y - 3 = -1/6 (x + 2)
6y - 18 = -x -2
x + 6y -16 = 0

[neutrino]

thanks for the help. Just wanted to verify one thing though, dont you muliply -1 by the (x + 2) making them both negative.

y - 3 = -1/6 (x + 2)
6y - 18 = -x -2
x + 6y -16 = 0
No, you don't.

(y-y0) = m(x-x0)

y0 = 3, x0 = -2, m = 1/6.