# Thread: Normals parallel to a line

1. ## Normals parallel to a line

My brother and I have been working on this problem for the better part of the day. I think we keep overlooking something. Please help us with this problem.

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

2. Originally Posted by AnDiesel
My brother and I have been working on this problem for the better part of the day. I think we keep overlooking something. Please help us with this problem.

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

$xy + 2x - y = 0$

$(x - 1)y = -2x$

$y = -\frac{2x}{x - 1}$

Now find the slope of the line tangent to this curve at (x, y):
$y' = \frac{2}{(x - 1)^2}$

The normal to the tangent line has the slope
$-\frac{1}{y'} = -\frac{(x - 1)^2}{2} = -\frac{1}{2}x^2 + x - \frac{1}{2}$

So given a point (x, y) on the original curve you now know the slope of the line normal to the curve at (x, y). I'll let you finish it out from here.

-Dan

3. Ok, I see what you did. I kept differentiating implicity and I must have been overlooking a mistake (even though it should have worked too). But I came up with y=-2x+3 , y=-2x-3 as the equations of my normals. Everything looks good when I graph it too, so I think this is right. Thank you very much for your quick responce and your help.

4. Originally Posted by topsquark
$xy {\color{red}+} 2x - y = 0$ Mr F says: Dang it!!

$(x - 1)y = 2x$

$y = \frac{2x}{x - 1}$

Now find the slope of the line tangent to this curve at (x, y):
$y' = \frac{-2}{(x - 1)^2}$

The normal to the tangent line has the slope
$-\frac{1}{y'} = -\frac{(x - 1)^2}{-2} = \frac{1}{2}x^2 - x + \frac{1}{2}$

So given a point (x, y) on the original curve you now know the slope of the line normal to the curve at (x, y). I'll let you finish it out from here.

-Dan
There's a small mistake (edited in red - dang it, the very first line. Topsquawk was obviously thinking about Dora at the time .... lol!!). There's obviously a flow-on effect but the corrections are simple.

You should get the x-coordinates of the required points on the given curve are x = 3 and x = -1.

5. Originally Posted by AnDiesel
Ok, I see what you did. I kept differentiating implicity and I must have been overlooking a mistake (even though it should have worked too). But I came up with y=-2x+3 , y=-2x-3 as the equations of my normals. Everything looks good when I graph it too, so I think this is right. Thank you very much for your quick responce and your help.
Obviously there's a bit more work if you use implicit differentiation (and therefore don't make y the subject). Anyway, since you mentioned it:

Implicit differentiation:

$y + x \frac{dy}{dx} + 2 - \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} \, (x - 1) + y + 2 = 0 \Rightarrow \frac{dy}{dx} = - \frac{(y +2)}{(x - 1)}$.

Therefore you require $-2 = \frac{x-1}{y+2} \Rightarrow -2y - 4 = x - 1$

$\Rightarrow x + 2y = -3$ ..... (1)

Since the points on the curve that the required normals pass through obviously lie on the curve, you also require

$xy + 2x - y = 0$ ...... (2)

Solve equations (1) and (2) simultaneously:

x = 3 and y = -3 or x = -1 and y = -1.

Equation of normal 1: m = -2 and passes through the point (3, -3). $y - (-3) = -2(x - 3) \Rightarrow y = -2x + 3$.

Equation of normal 2: m = -2 and passes through the point (-1, -1). $y - (-1) = -2(x - (-1)) \Rightarrow y = -2x - 3$.

6. Thank you for your replies; they were a great help to me.

7. Originally Posted by mr fantastic
There's a small mistake (edited in red - dang it, the very first line. Topsquawk was obviously thinking about Dora at the time .... lol!!). There's obviously a flow-on effect but the corrections are simple.

You should get the x-coordinates of the required points on the given curve are x = 3 and x = -1.

Thanks for pointing out the mistake. (Not for the Dora comment!) I have corrected my original post.

-Dan