# Thread: Normal line, parallel lines

1. ## Normal line, parallel lines

find an equation of the normal line to y=x2-5x+4 that is parallel to the line x-3y=5.

2. ## Re: Normal line, parallel lines

One way to work this problem:

Implicitly differentiate the parabola with respect to y and solve for $-\frac{dx}{dy}$. Equate this to the slope of the given line, allowing you to find the point of intersection of the normal line and the parabola. Then use the point-slope formula to find the equation of the normal line.

3. ## Re: Normal line, parallel lines

What is the slope of the line x- 3y= 5? Since you want a line parallel to that, you want a line with that same slope.

What is the derivative of $x^2- 5x+ 4$ for general x? That will be the slope, m, of the tangent line to the curve. A normal line will have slope -1/m. For what x is the derivative equal to -1 over the slope from the first question?

4. ## Re: Normal line, parallel lines

the derivative of the the first equation is y'=2x-5 and the derivative of the second equation is y'=1/3 so now what do i do?
Originally Posted by HallsofIvy
What is the slope of the line x- 3y= 5? Since you want a line parallel to that, you want a line with that same slope.

What is the derivative of $x^2- 5x+ 4$ for general x? That will be the slope, m, of the tangent line to the curve. A normal line will have slope -1/m. For what x is the derivative equal to -1 over the slope from the first question?

5. ## Re: Normal line, parallel lines

You don't need the derivative of the second line in this problem. First put the second line in y = mx + b form. It has a slope of $1/3$, a line normal to that function would have a slope that's the negative reciprocal $-3$. The negative reciprocal or -1/m of 1/3 is -3. To answer HallofIvy's question,
For what x is the derivative equal to -1 over the slope from the first question?
we can equate the negative reciprocal of the second line $-3$ with the derivative of the first function (which is what MarkFL2 was saying). $-3 = 2x -5$, this gives $x = 1$. So, if you plug 1 into the derivative of the parabola you'll get $-3$ which is the negative reciprocal of the second line's slope. Plug 1 into the equation of the parabola and we'll get y = 0, so the point where the normal line intersects the parabola is (1, 0), and using the point slope form with the exact same slope of the second line $(y-0) = 1/3(x-1)$ gives $y = 1/3x -1/3$.

I was working on a similar problem and I checked this site out here for help: Find the equation of the normal line of the given parabola that is parallel to a line

,

,

,

,

,

,

# normals and lines parallel

Click on a term to search for related topics.