# Finding a point of tangency on a curve (1 Viewer)

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#### NJJ289

1. The problem statement, all variables and given/known data

A line Y exists which is tangent to the curve y=x^2 at some point(s) and which passes through the point (2,0)

2. Relevant equations

3. The attempt at a solution

dy/dx=2x

Y(2)=0=(2x)(2)+b
b=-4x (x being the first coordinate of the points I'm trying to find)

I know (0,0) is one of the points but I don't know how to find it or the other point algebraically.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### berkeman

Mentor
1. The problem statement, all variables and given/known data

A line Y exists which is tangent to the curve y=x^2 at some point(s) and which passes through the point (2,0)

2. Relevant equations

3. The attempt at a solution

dy/dx=2x

Y(2)=0=(2x)(2)+b
b=-4x (x being the first coordinate of the points I'm trying to find)

I know (0,0) is one of the points but I don't know how to find it or the other point algebraically.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Welcome to the PF.

The problem statement seems to be missing something, or the equation needs more flexibility. The point (2,0) does not lie on the curve y=x^2.

EDIT -- Oh wait, I see now. The line passes through (2,0) and is also tangent to the parabola at some other point. Let me look at it again...

#### NJJ289

yeah I made the mistake of assuming that the line actually passed through the point of tangency. I tried m=y-Y/x-X which yielded a very incorrect answer.

#### berkeman

Mentor
I think you are really close. You have the slope of the line correct, 2x at x=2, so the slope is 4. Since that is the slope and the line passes through (2,0), you can find the y-intercept b directly..

#### berkeman

Mentor
I think you are really close. You have the slope of the line correct, 2x at x=2, so the slope is 4. Since that is the slope and the line passes through (2,0), you can find the y-intercept b directly..
Oh wait, that may not be right...

#### berkeman

Mentor

You already know that at the point of tangency, dy/dx = 2x.

And the equation for the straight line is y = mx + b

But m is dy/dx at the point of tangency, so make that substitution into y=mx + b and what do you get for the equation? Then you know that the point (2,0) has to lie on the line, so plug that point into the equation to find b.

Then you have two equations y=x^2 and y=mx+b that you should be able to use to find where the line is tangent to the parabola....

Does that work?

#### NJJ289

I guess the problem is that there are different X values here. If you do that you end up with y=0=(2X)(2)+b
and then a seperate function Y=X^2. I can solve for b in terms of X, X in terms of Y... but not get the answer. There's some step missing...

#### berkeman

Mentor
I guess the problem is that there are different X values here. If you do that you end up with y=0=(2X)(2)+b
and then a seperate function Y=X^2. I can solve for b in terms of X, X in terms of Y... but not get the answer. There's some step missing...
Where you wrote y=0=(2X)(2)+b

you forgot to substitute for that last X in the equation. What is b = ?

#### NJJ289

What am I supposed to substitute for X? I don't think X=x so it wouldn't be valid to say b=-8 would it?

#### berkeman

Mentor

What am I supposed to substitute for X? I don't think X=x so it wouldn't be valid to say b=-8 would it?
Yes, I got b = -8.

We may have gotten it the same way, but I wrote it a bit differently:

dy/dx = 2x

y = mx + b = (2x)(x) + b = 2x^2 + b

Substitute the point (2,0) which is on that line:

0 = 8 + b, so b = -8

Now you can write out the equation y = mx + b (substitute m = 2x and b = -8), which is one equation, and the other equation is y = x^2 for the parabola. You should be able to solve those two equations for the point where the line and parabola meet...

#### NJJ289

So something like this?:

0=2x^2-8, 8/2=x^2, 4=x^2, x= +2 or -2

-2 =rad(y) is nonreal, so 2= rad(y), y=4

x=2, y=4.

Problem is:

equating the tangent line (y) with the parabola (Y) is impossible... y never actually touches Y. They should share no points.

#### berkeman

Mentor
So something like this?:

0=2x^2-8, 8/2=x^2, 4=x^2, x= +2 or -2

-2 =rad(y) is nonreal, so 2= rad(y), y=4

x=2, y=4.

Problem is:

equating the tangent line (y) with the parabola (Y) is impossible... y never actually touches Y. They should share no points.
I'm not following what you are writing, but my impression from the problem statement...

A line Y exists which is tangent to the curve y=x^2 at some point(s) and which passes through the point (2,0)
...is that the line is tangent to the parabola at a point (means it touches it), and also passes through (2,0).

Don't substitute the point (2,0) into anything until you write the two equations and then get them down to one equation. Once you have the one equation, substituting (2,0) should get you the value of y where the tangent line touches the parabola, and then you can go from there to find the x value of the intercept as well...

#### NJJ289

wait so if it's tangent it WILL touch a point on the graph? Yikes! that helps a lot... I was having a lot of trouble because I thought tangent lines get really, really close but don't actually touch the curve.

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