- Thread starter
- #1

#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

i know how to calculate normal line on point but i dont know how to do this problem.

"find an equation of the normal line o the parabola y=x^2-5x+4 tha is parallel to the line x-3y=5."

- Thread starter Petrus
- Start date

- Thread starter
- #1

- Feb 21, 2013

- 739

i know how to calculate normal line on point but i dont know how to do this problem.

"find an equation of the normal line o the parabola y=x^2-5x+4 tha is parallel to the line x-3y=5."

- Moderator
- #2

- Feb 7, 2012

- 2,707

If the normal has slope $1/3$ then the tangent has slope $-3$. Can you find a point on the parabola where the tangent has slope $-3$?

i know how to calculate normal line on point but i dont know how to do this problem.

"find an equation of the normal line o the parabola y=x^2-5x+4 tha is parallel to the line x-3y=5."

- Thread starter
- #3

- Feb 21, 2013

- 739

Hello,If the normal has slope $1/3$ then the tangent has slope $-3$. Can you find a point on the parabola where the tangent has slope $-3$?

I dont understand where you get them all, could you possible tell me how you know that?

- Moderator
- #4

- Feb 7, 2012

- 2,707

You are told that the normal line is parallel to the line $x-3y=5$, which you can write as $y = \frac13x-\frac53.$ That is a line with slope $\frac13.$ The tangent is perpendicular to the normal, and you should know that if a line has slope $m$, then the perpendicular line has slope $-1/m.$ So if the normal has slope $\frac13$ then the tangent has slope $-3$. Does that help?could you possible tell me how you know that?

- Thread starter
- #5

- Feb 21, 2013

- 739

If I know that how shall I proceed? I can't make any progress.

Last edited by a moderator:

- Admin
- #6

If you compute the derivative of the parabola $f(x)=x^2-5x+4$, equate that to -3 and solve for $x$, you will then be able to determine the point on the parabola at which the normal line will intersect it.

Then you will have determined a point $(x,f(x))$ and the slope $m=-3$, and you may then apply the point-slope formula to obtain the equation of the desired normal line.

Can you follow these steps, and show us your work?

- Thread starter
- #7

- Feb 21, 2013

- 739

Hello Mark,

If you compute the derivative of the parabola $f(x)=x^2-5x+4$, equate that to -3 and solve for $x$, you will then be able to determine the point on the parabola at which the normal line will intersect it.

Then you will have determined a point $(x,f(x))$ and the slope $m=-3$, and you may then apply the point-slope formula to obtain the equation of the desired normal line.

Can you follow these steps, and show us your work?

we start with calculate x

first i derivate the function $f(x)=x^2-5x+4$ and get $f'(x)=2x-5$ and then i set that =-3 and get $x=1$ and then i put $x=1$ on orginal function and get $y=0$ so we get the point slope formula $y-y1=m(x-x1)$ and get $y-0=-3(x-1)$ and then $y=3x-3$

- Admin
- #8

- Thread starter
- #9

- Feb 21, 2013

- 739

Hello Mark,I agree with your work until the very last step. You correctly wrote:

$y-0=-3(x-1)$

Now, try distributing the -3 on the right again...

Sorry i forgot i had -3 i mean $y=-3x+3$

- Admin
- #10

- Thread starter
- #11

- Feb 21, 2013

- 739

Thank you Mark and the one who replied early! Now i understand how to do it thanks to the help! Have a nice day!Yes!

Good work, Petrus! I appreciate that you read the suggestions, applied them and posted your work as a means of following through! This lets us know that you understood the suggestions.

- Moderator
- #12

- Feb 7, 2012

- 2,707

Good work!$y=-3x+3$

- Admin
- #13