# Calculus

posted by
**Jen**
.

Find normals to the curve xy+2x-y=0 that are parallel to the line 2x+y=0

I have the answer: at(-1,-1), y=-2x-3, and at (3,-3), y=-2x+3

How do we get this? Thanks.

xy + 2x - y=0

x dy/dx + y + 2 - dy/dx=0

dy/dx (x-1)= -y-2

dy/dx= - (y+2)/(x-1)

The normal to this has a slope of

(x-1)/(y+2)

but this has to be parallel to a slope of -2

(x-1)/(y+2) = -2

x-1= -2y -4

x= -(2y +3 ) and the x,y has to be on the

curve xy + 2x - y=0 ,or

(-(2y +3 )y) + 2(-(2y +3 )) - y=0

-2y^2-3y -4y-6-y=0

2y^2 + 8y + 6=0

y^2 + 4y +3=0

(y+3)(y+1)=0

and you have your two y solutions, put these back into the original equation to get the x values.

Sorry about the transcription error in the first post. Thanks for catching it.