# Finding a point of tangency on a curve

• NJJ289
In summary, the line Y is tangent to the parabola at a point (means it touches it), and also passes through (2,0).
NJJ289

## Homework 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)

## 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.

NJJ289 said:

## Homework 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)

## 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.

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...

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.

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 said:
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...

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?

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 separate 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...

NJJ289 said:
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 separate 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 = ?

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?

NJJ289 said:

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...

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.

NJJ289 said:
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...

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.

## 1. How do you find the point of tangency on a curve?

To find the point of tangency on a curve, you will need to use calculus and the derivative of the curve. Set the derivative equal to the slope of the line tangent to the curve at a specific point, and then solve for the x-value. This x-value will be the x-coordinate of the point of tangency. Plug this value back into the original equation to find the y-coordinate.

## 2. What information do you need to find the point of tangency on a curve?

To find the point of tangency on a curve, you will need the equation of the curve and its derivative. You will also need the x-coordinate of the point where you want to find the point of tangency.

## 3. Can there be more than one point of tangency on a curve?

Yes, there can be more than one point of tangency on a curve. This usually occurs when the curve has multiple "hills and valleys" or when the slope of the curve changes drastically at different points.

## 4. How do you know if a point is a point of tangency on a curve?

A point is a point of tangency on a curve if it lies on the curve and if the slope of the curve at that point is equal to the slope of the line tangent to the curve at that point. This can be determined by setting the derivative of the curve equal to the slope of the tangent line and solving for the x-coordinate of the point.

## 5. Can you use a graphing calculator to find the point of tangency on a curve?

Yes, a graphing calculator can be used to find the point of tangency on a curve. Most graphing calculators have a "zero" or "intersect" function that allows you to find the x-coordinate of the point of tangency by setting the derivative equal to the slope of the tangent line. You can then plug this value back into the original equation to find the y-coordinate.

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