# Homework Help: Normal line question

1. Apr 11, 2007

### 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

2. Apr 12, 2007

### neutrino

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?

3. Apr 12, 2007

### mathmann

Is it the reciprocal?

1
_ ?

6

4. Apr 12, 2007

### Tom Mattson

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

5. Apr 12, 2007

### mathmann

its not the reciprocal?

my current answer in standard form is..

x + 6y - 16 = 0

Last edited: Apr 12, 2007
6. Apr 12, 2007

### daniel_i_l

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

7. Apr 12, 2007

### mathmann

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

Is it correct now?

8. Apr 12, 2007

### neutrino

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

9. Apr 12, 2007

### 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

10. Apr 12, 2007

### neutrino

No, you don't.

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

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