Find Equation of Normal Line to y=x^2-5x+4 Parallel to x-3y=5

In summary: We want to help you understand mathematics better, so that you can be successful in your studies.In summary, the conversation discusses finding the equation of a normal line on a parabola that is parallel to a given line. The expert suggests finding a point on the parabola where the tangent has a specific slope, and then using the point-slope formula to find the equation of the normal line. The person asking the question follows these steps and successfully finds the equation of the normal line. The expert and others express their satisfaction with the outcome and the person's improved understanding.
  • #1
Petrus
702
0
Hello,
i know how to calculate normal line on point but i don't 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."
 
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  • #2
Petrus said:
Hello,
i know how to calculate normal line on point but i don't 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 $-3$?
 
  • #3
Opalg said:
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 don't understand where you get them all, could you possible tell me how you know that?
 
  • #4
Petrus said:
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?
 
  • #5
If I know that how shall I proceed? I can't make any progress.
 
Last edited by a moderator:
  • #6
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?
 
  • #7
MarkFL said:
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 $y=3x-3$
 
  • #8
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...
 
  • #9
MarkFL said:
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 $y=-3x+3$
 
  • #10
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 let's us know that you understood the suggestions. :cool:
 
  • #11
MarkFL said:
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 let's 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!
 
  • #12
Petrus said:
$y=-3x+3$
Good work! (Muscle) (Star)
 
  • #13
We are glad to help, and it is gratifying to hear that your understanding has increased! (Clapping)

This is our goal.
 

FAQ: Find Equation of Normal Line to y=x^2-5x+4 Parallel to x-3y=5

How do I find the equation of a normal line to a given curve?

To find the equation of a normal line to a curve, we need to follow these steps:
1. Find the derivative of the given curve.
2. Substitute the x-value of the point of tangency into the derivative to find the slope of the tangent line.
3. Take the negative reciprocal of this slope to find the slope of the normal line.
4. Use the point of tangency and the slope of the normal line to write the equation of the normal line in point-slope form.
5. Simplify the equation if necessary to obtain the final equation of the normal line.

What is a normal line?

A normal line is a line that is perpendicular to a given curve at a specific point of tangency. This means that the slope of the normal line is the negative reciprocal of the slope of the tangent line at that point.

Can a normal line be parallel to a given line?

Yes, a normal line can be parallel to a given line. This occurs when the given line is also perpendicular to the curve at the point of tangency. In other words, the slope of the given line is the negative reciprocal of the slope of the tangent line at that point.

How do I determine if a normal line is parallel to a given line?

To determine if a normal line is parallel to a given line, we need to compare the slopes of the two lines. If the slope of the normal line is equal to the slope of the given line, then they are parallel. Remember, the slope of the normal line is the negative reciprocal of the slope of the tangent line at the point of tangency.

Can I find the equation of a normal line without finding the derivative?

No, the derivative is necessary in finding the equation of a normal line. This is because the derivative gives us the slope of the tangent line, which is needed to find the slope of the normal line. Without the derivative, we would not have enough information to determine the equation of the normal line.

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