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Normal line

Petrus

Well-known member
Feb 21, 2013
739
Hello,
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."
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Hello,
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."
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 $-3$?
 

Petrus

Well-known member
Feb 21, 2013
739
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 $-3$?
Hello,
I dont understand where you get them all, could you possible tell me how you know that?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
could you possible tell me how you know that?
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?
 

Petrus

Well-known member
Feb 21, 2013
739
If I know that how shall I proceed? I can't make any progress.
 
Last edited by a moderator:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Petrus,

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?
 

Petrus

Well-known member
Feb 21, 2013
739
Hello Petrus,

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?
Hello Mark,
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 $y=3x-3$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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...
 

Petrus

Well-known member
Feb 21, 2013
739
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...
Hello Mark,
Sorry i forgot i had -3 i mean $y=-3x+3$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes! (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. :cool:
 

Petrus

Well-known member
Feb 21, 2013
739
Yes! (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. :cool:
Thank you Mark and the one who replied early!:) Now i understand how to do it thanks to the help!:) Have a nice day!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We are glad to help, and it is gratifying to hear that your understanding has increased! (Clapping)

This is our goal.