NationStates Jolt Archive

I need help. I'm kind of stuck...

Here is the problem:

Prove that for any point (a,b) not equal to (0,0) on the parabola y=x^2, the normal line intersects the graph twice.


Here is what I have so far:

y = x^2, therefore y' = 2x
I also rewrote y = x^2 as b = a^2, and said that x = a

Because the derivative is 2x, the slope of the tangent line is 2x, therefore making the inverse (the normal line) -1/2x
You would find the actual slope by plugging the x value of the parabola at the point you wish to find the slope of into that mini-function.

Anyway, next, to avoid confusion in the next steps, I rewrote the slope as -1/2a, simply replacing the x with an a.
To find the equation for the line at point (a,b), you must use the following equation:
y-b = m(x-a)

We already know that m = -1/2a, and that b = a^2, and that y = x^2, so I rewrote that equation (after some algebra) as:
0 = -(x-a)/2a + (a^2) - (x^2)

This is where I am at right now. Am I on the right track and just need to keep doing algerba until I show that there are two values for a that can work in the equation? Or did I mess up somewhere and I am completely off track? Your help is appreciated, thanks.

I need help. I'm kind of stuck...

Here is the problem:

Prove that for any point (a,b) not equal to (0,0) on the parabola y=x^2, the normal line intersects the graph twice.


Here is what I have so far:

y = x^2, therefore y' = 2x
I also rewrote y = x^2 as b = a^2, and said that x = a

Because the derivative is 2x, the slope of the tangent line is 2x, therefore making the inverse (the normal line) -1/2x
You would find the actual slope by plugging the x value of the parabola at the point you wish to find the slope of into that mini-function.

Anyway, next, to avoid confusion in the next steps, I rewrote the slope as -1/2a, simply replacing the x with an a.
To find the equation for the line at point (a,b), you must use the following equation:
y-b = m(x-a)

We already know that m = -1/2a, and that b = a^2, and that y = x^2, so I rewrote that equation (after some algebra) as:
0 = -(x-a)/2a + (a^2) - (x^2)

This is where I am at right now. Am I on the right track and just need to keep doing algerba until I show that there are two values for a that can work in the equation? Or did I mess up somewhere and I am completely off track? Your help is appreciated, thanks.
since tou statet that x=a, the equasion is totally useless
0 = 0/2a + 0

I need help. I'm kind of stuck...

Here is the problem:

Prove that for any point (a,b) not equal to (0,0) on the parabola y=x^2, the normal line intersects the graph twice.


Here is what I have so far:

y = x^2, therefore y' = 2x
I also rewrote y = x^2 as b = a^2, and said that x = a

Because the derivative is 2x, the slope of the tangent line is 2x, therefore making the inverse (the normal line) -1/2x
You would find the actual slope by plugging the x value of the parabola at the point you wish to find the slope of into that mini-function.

Anyway, next, to avoid confusion in the next steps, I rewrote the slope as -1/2a, simply replacing the x with an a.
To find the equation for the line at point (a,b), you must use the following equation:
y-b = m(x-a)

We already know that m = -1/2a, and that b = a^2, and that y = x^2, so I rewrote that equation (after some algebra) as:
0 = -(x-a)/2a + (a^2) - (x^2)

This is where I am at right now. Am I on the right track and just need to keep doing algerba until I show that there are two values for a that can work in the equation? Or did I mess up somewhere and I am completely off track? Your help is appreciated, thanks.

The mistake is here:

y=x^2 becomes y=+or- sqr(x) [plus or minus the square root of x.]

THe proof is right before you... + or -. For any value of X other than 0, there are TWO values for y. Therefore, two points of intersection. *nod*

since tou statet that x=a, the equasion is totally useless
0 = 0/2a + 0
No. The a becomes a constant while x remains a variable. I don't think you've had any calculus or else you'd understand exactly what I'm doing.

The mistake is here:

y=x^2 becomes y=+or- sqr(x) [plus or minus the square root of x.]

THe proof is right before you... + or -. For any value of X other than 0, there are TWO values for y. Therefore, two points of intersection. *nod*
Uh, that's not why there are two intersections. That's why a parabola looks like a parabola on both sides of the y-axis. There is a y value for a +x and the exact same value on the -x. It's why when y=4, x can be +2 or -2, but it doesn't prove that the normal line intersects the parabola twice.

It'd be nice if someone who has had some college calculus could help me.

Well... I took calculus but didn't really like it. But I think my one question is why you took the slope and substituted "a" in for "x"? Why doesn't it stay as (-x/2)?

If you don't substitute, you get a nice equation, 3x^2 -ax -2a^2. Hehe.. .manipulating math is fun!

I'm basically clueless here. :)

No. The a becomes a constant while x remains a variable. I don't think you've had any calculus or else you'd understand exactly what I'm doing.
Dude, I'm doing Math you've never even heard of.

Try proving that at a certain point
a*x > b*x
where a*x is the derivative (don't mention my term use, I'm dutch)
and b*x is the normal line

Uh, that's not why there are two intersections. That's why a parabola looks like a parabola on both sides of the y-axis. There is a y value for a +x and the exact same value on the -x. It's why when y=4, x can be +2 or -2, but it doesn't prove that the normal line intersects the parabola twice.

I tackled it backwards. But you just explained it yourself. There are two values of x that yield the same value for y. WHich means that for any and all values of y, there are two different values of x. (2,4) and (-2,4) for instance.

The important detail in your equation is to note that the removal of the ^2(which sooner or later must come out) will create two separate equations. One with a +sqr, and one with a -sqr.

Because the slope is -1/2x, but the x for the slope isn't the same x in the new equation...

For instance, the slope of the line tangent to the graph y = x^2 at x = 2...
Derive the equation:y' = 2x
x = 2, therefore y' = m = 4
The graph of the tangent line has a slope of 4, and to find the equation of that line, (y-4) = 4(x-2)

The y-4 represents the y point that is being solved for subtracted from y
The x-2 represents the x point that is being solved for subtracted from x
The y point and the x point in the above problem (the first one I placed) were replaced by (a,a^2) as opposed to (2,4) since we are solving theoritically, as a proof.
Anyway, the equation of that line comes out to be y = 4x+4. That's the tangent line.
The normal line (perpandicular to the tangent line) is y = (-1/4)x + (9/2). To find the other intersection, you replace y with x^2 (since you know y = x^2 and that there are two intersections). After a lot of algebra, you come up with (4x+9)(x-2) = 0, which gives you your two Xs, 2 and -9/4. You simple plug those numbers into y = x^2 to solve for the Ys and find the two points of intersection.

However, this is not the problem. The problem is solving this theoritically to serve as a proof, everything in this post is just an explanation so you guys are more clear on what the problem is actually asking for...

Any one?

Dude, I'm doing Math you've never even heard of.

Try proving that at a certain point
a*x > b*x
where a*x is the derivative (don't mention my term use, I'm dutch)
and b*x is the normal line
Do you even read the posts?

Do you even read the posts?
I thought you were trying to impress me with the type of math you are doing...because I didn't understand that post--and still don't.

...so feel free to explain.

It's really hard for me to explain because I'm not into your terminology.

You calculated the normal line on a certain point on the parabola right?

Any straight line makes:
1) no touch with the parabola, no the case since the normal line always intersects the parabola
2) 1 touch, but the parabola makes a 90 degree angle with the parabola so it intersects
3) 2 intersections, because at some point the "angle" (when y=x^2 then y'=2x and 2 is the "angle", I don't know how to call it) of the parabola will be greater that the one from the normal line (because the parabola "angle" increases/decreases, because it's a parabola) so the normal line has to intersect twice or have an infinite "angle", which is not possible in calculus

is it of any help?

It's really hard for me to explain because I'm not into your terminology.

You calculated the normal line on a certain point on the parabola right?

Any straight line makes:
1) no touch with the parabola, no the case since the normal line always intersects the parabola
2) 1 touch, but the parabola makes a 90 degree angle with the parabola so it intersects
3) 2 intersections, because at some point the "angle" (when y=x^2 then y'=2x and 2 is the "angle", I don't know how to call it) of the parabola will be greater that the one from the normal line (because the parabola "angle" increases/decreases, because it's a parabola) so the normal line has to intersect twice or have an infinite "angle", which is not possible in calculus

If I want to get credit, I need an analytical proof. Besides, a line can touch a parabola only once and not be tangent. Try y = x^2 and x = 2. The line x = 2 crosses the parabola at (2,4) and no other place, yet it is not tangent to the parabola.

If I want to get credit, I need an analytical proof. Besides, a line can touch a parabola only once and not be tangent. Try y = x^2 and x = 2. The line x = 2 crosses the parabola at (2,4) and no other place, yet it is not tangent to the parabola.
it's on a random point (a,b) on the parabola and the normal line is ALWAYS on a 90 degree angle with the parabola

Hey, Opal Isle my 8th grade daughter (Suicidal Librarians) printed out your problem for me because I teach Advanced Placement Calculus at my high school. You have actually done a nice job getting this problem set up.

Here's what I did. I used the slope of the normal (-1/2a) and the point on the curve (a, a^2) and the point-slope form to get the equation of the normal line below (note that I have solved it for y):

y = (-1/2a)x + 1/2 + a^2

Now, you want to find where this line intersects y = x^2 so set the equations equal to one another:

x^2 = (-1/2a)x + 1/2 + a^2 then move everything to one side to get

it equal to zero.

x^2 + (1/2a)x -1/2 -a^2 = 0

Next, I multiplied both sides by 2a to clear the fraction.

2a*x^2 + x - a - 2a^3 = 0

Now, use the quadratic formula. This can get confusing since
a was used in the original problem, so I am going to use capital letters
for the A, B, and C in the quadratic formula.

In this case, A = 2a, B = 1, and C = -a - 2a^3.

After substituting these values and doing some simplification you
will have x = (-1 plus or minus the sqr. root of (1 + 8a^2 + 16a^4))/(4a).

This will become x = (-1 plus or minus the sqr. root of ((1 + 4a^2)^2))/(4a).

This becomes x = (-1 plus or minus (1 + 4a^2))/(4a).

From here, the two solutions are x = a and x = (-1 - 2a^2)/(2a).

Therefore, two points of intersection exist and the x-coordinates of these points of intersection are listed above.

I hope you can follow this, it is tough to explain these things without being
able to use all of the proper math symbols.

Yeah, that really was my dad. ^

I'm not that smart.

The mistake is here:

y=x^2 becomes y=+or- sqr(x) [plus or minus the square root of x.]

THe proof is right before you... + or -. For any value of X other than 0, there are TWO values for y. Therefore, two points of intersection. *nod*
Yep, he is right.
That's what happens when you take precalc twice. You still don't understand how it works, you just remember the process.

Hey, Opal Isle my 8th grade daughter (Suicidal Librarians) printed out your problem for me because I teach Advanced Placement Calculus at my high school. You have actually done a nice job getting this problem set up.

Here's what I did. I used the slope of the normal (-1/2a) and the point on the curve (a, a^2) and the point-slope form to get the equation of the normal line below (note that I have solved it for y):

y = (-1/2a)x + 1/2 + a^2

Now, you want to find where this line intersects y = x^2 so set the equations equal to one another:

x^2 = (-1/2a)x + 1/2 + a^2 then move everything to one side to get

it equal to zero.

x^2 + (1/2a)x -1/2 -a^2 = 0

Next, I multiplied both sides by 2a to clear the fraction.

2a*x^2 + x - a - 2a^3 = 0

Now, use the quadratic formula. This can get confusing since
a was used in the original problem, so I am going to use capital letters
for the A, B, and C in the quadratic formula.

In this case, A = 2a, B = 1, and C = -a - 2a^3.

After substituting these values and doing some simplification you
will have x = (-1 plus or minus the sqr. root of (1 + 8a^2 + 16a^4))/(4a).

This will become x = (-1 plus or minus the sqr. root of ((1 + 4a^2)^2))/(4a).

This becomes x = (-1 plus or minus (1 + 4a^2))/(4a).

From here, the two solutions are x = a and x = (-1 - 2a^2)/(2a).

Therefore, two points of intersection exist and the x-coordinates of these points of intersection are listed above.

I hope you can follow this, it is tough to explain these things without being
able to use all of the proper math symbols.

Tell your dad I said thanks. Heh, when I was working it I saw the a^3 and was kinda iffy about doing it because it's been like two years since I learned the quadratic thing...time to refresh..

Actually...I don't understand this step:
This becomes x = (-1 plus or minus (1 + 4a^2))/(4a).

From here, the two solutions are x = a and x = (-1 - 2a^2)/(2a).

So if it's possible, I'd appreciate a further explanation of the last step by anyone who understands it. (I understand the x =a)

Quadratic formula.
If ax^2+bx+c=0, then
x= (-b +/- sqrt (b^2-4ac))/2a
http://planetmath.org/encyclopedia/DerivationOfQuadraticFormula.html

Quadratic formula.
If ax^2+bx+c=0, then
x= (-b +/- sqrt (b^2-4ac))/2a
http://planetmath.org/encyclopedia/DerivationOfQuadraticFormula.html
Sorry, I don't really consider that an explanation. It doesn't make sense to me. How do you get from
x = (-1 plus or minus (1 + 4a^2))/(4a)
to
x = (-1 - 2a^2)/(2a)

What happens to the plus or minus? Plus, I don't see how 2a^2 goes to 4a^2 or how 4a goes to 2a

The second solution comes from the "minus" part of the quadratic formula:
x = (-1 - (1 +4a^2))/(4a). If you subtract the stuff in parentheses it becomes (-2 - 4a^2)/(4a). Now just factor a 2 out of the numerator.
(2(-1 - 2a^2))/(4a). Then cancel it with a 2 in the denominator and you get the 2nd solution that I provided earlier:

x = (-1 - 2a^2)/(2a)

Ah...I see. And had you done the plus part of it all, it'd all reduce down to a/1 or a. Okay. I'm still refreshing the quadratic formula stuff, but I think I'm getting it. Thanks a lot.

Glad I could help--sorry about the leap to the final solutions earlier.

I got a kick out of reading some of the early responses you were getting to your problem. Intuitively, it is obvious that a normal line will intersect twice (except the one through the origin), but that just doesn't cut it as a proof!!

Glad I could help--sorry about the leap to the final solutions earlier.

I got a kick out of reading some of the early responses you were getting to your problem. Intuitively, it is obvious that a normal line will intersect twice (except the one through the origin), but that just doesn't cut it as a proof!!
Exactly...and with a D in my honors calc class, I don't think that just using intuition (especially on group projects) is going to cut it. Anyway...I get this stuff, I just don't get it all the way through. Mid term is a week from Thursday...

Oh well, my mom is about to complete her masters in Civil Engineering and said she always had Cs and Ds in her math classes, but hey, I have an A in my Honors Comp class.

Actually, this stuff isn't really hard math when I think about it. The hard part is pooling all the math skills all together for one problem. I know all these skills, just can't remember them and can't intuitively select the specific formulae needed for a particular problem at the particular moment.