What Are the Tangent Lines to the Parabola y = x² Through (-1, 3)?

[jmed]

Homework Statement

Give the equation of the two lines through the point (-1, 3) that are tangent to the parabola y= x^2

Homework Equations

The Attempt at a Solution

 

[ystael]

What have you done so far?

Also, are you sure you copied the question correctly?

 

[jmed]

i really do not no where to start?? I did copy the question correctly.

 

[jmed]

Equation of a line

Homework Statement

Give the equation of the two lines through the point (-1, 3) that are tangent to the parabola y= x^2

Homework Equations

The Attempt at a Solution

Not sure where to start?? Do I find the deriviative of x^2? I know that equals 2x...

 

[rock.freak667]

So you know that dy/dx gives the gradient of the tangent at a point (x,y)

dy/dx=2x.

So what is the gradient of the tangent at the point (-1,3)

 

[Mark44]

Not sure where to start??

Start by drawing a graph..

 

[jmed]

I did draw the graph...still not helping...So if dy/dx is equal to 2x, would you plug -1 in for x?

 

[Mark44]

No, not at all.

Looking at your graph, about where do the lines through (-1, 3) hit the parabola so that they are tangent to it?

 

[jmed]

well there is only one line that goes through it...and I don't know where it is tangent? it is negative...

 

[Mark44]

I see two lines. When you say "it is negative" please be more more specific. What is negative?

 

[jmed]

it is negative as x approaches zero?

 

[Mark44]

What is "it"? Please be more specific. I have no idea what you are trying to say.

 

[jmed]

the tangent line on the parabola at point (-1,3) as it approaches zero

 

[Mark44]

How can a line be negative? In "as it approaches zero" what is it? Please stop using pronouns.

The point (-1, 3) is NOT on the parabola!

 

[sponsoredwalk]

You obviously don't know what you're being asked to find. That's fine, those questions used to throw me off too...

This video will show you an example of what you're actually being asked to do, it will also give you a better understanding of why you're being asked to find this line and also why the derivative is actually being asked of you...

http://www.5min.com/Video/The-Equation-of-a-Tangent-Line-169041763

I advise watching as many videos on this site, made by this guy as possible. For calculus, they will give you an understanding of what you are doing & that's a great thing if your pre-calculus is weak.

Also, this video will only give you one of the lines you're looking for, once you have fully watched this video (and the related ones if you can!) then I want you to find out what the "Normal" to a line. This is also called a Perpendicular line. This is what you're being asked to find but they don't explicitly state it as you're supposed to know this, but don't fret, just learn it now :)btw, the point (-1, 3) is not on the parabola, the point (-1,1) is, then (-2,4), maybe you wrote the wrong thing down here...

 

[Mark44]

btw, the point (-1, 3) is not on the parabola, the point (-1,1) is, then (-2,4), maybe you wrote the wrong thing down here...

Or not. AFAIK the OP wrote the problem correctly. It doesn't matter that (-1, 3) isn't on the parabola. The goal is to find two lines through this point that are tangent to the parabola. I'm trying to get the OP to at least have a visual understanding of what's going on in the problem.

I'm not being too successful, so far.

 

[jmed]

I still do not understand why there is two equations?? I know the derivative is 2x... so that means the slope = 2...the point (-1,1) is a pt. on the parabola. So, could I use this for point slope form?

 

[Mark44]

Is the line from (-1, 3) to (-1, 1) tangent to the parabola? You can ignore what sponsoredwalk said about normal or perpendicular lines. This problem has nothing to do with them.

Do you have a graph of the parabola and the point (-1, 3)? This point is inside the parabola. One of the tangent lines would have to be very steep, and with a negative slope. Can you show this on your graph?

I think it will be extremely difficult or even impossible for you to solve this problem without having a clear image of what you are trying to do.

 

[jmed]

i see that (-1,3) is inside the parabola...I just don't understand how the line passes through that point while being tangent to the parabola?? Wouldn't that line intersect the parabola then?

 

[ystael]

The problem is that the function $$f(x) = x^2$$ is what's called a convex function, which means (among other things) that the graph of $$f$$ lies above all its tangent lines and below all its chords. You can think of a convex function as one that "bends upward" everywhere; you know $$f$$ is convex because the second derivative of $$f$$ is the constant function $$2$$, which is everywhere positive. But the point $$(-1, 3)$$ lies above the graph of $$f$$, since $$f(-1) = 1 < 3$$. This means that $$(-1, 3)$$ does not lie on any tangent line to the graph of $$f$$.

 

[jmed]

so how do I find two equations of the ?LINES? that pass through the point (-1,3)?

 

[jmed]

So am I finding a line that is parallel to the line tangent to the parabola at point (-1,3)?

 

[jmed]

So, since the equation of the tangent line = y-3=2(x-(-1))...the normal line (-1,3) would equal y-3=-1/2(x-(-1))? correct??

 

[Mark44]

So am I finding a line that is parallel to the line tangent to the parabola at point (-1,3)?

No, because (-1, 3) is not a point on the parabola.

After thinking about this for a while, I've come to the conclusion that your earlier comments about not being able to find any tangent lines were correct. IOW, it's not possible to find a line through (-1, 3) that is tangent to the graph of y = x². I apologize for wasting your time with bad advice.

Are you sure that you have posted the problem correctly?

 

[HallsofIvy]

No, because (-1, 3) is not a point on the parabola.

After thinking about this for a while, I've come to the conclusion that your earlier comments about not being able to find any tangent lines were correct. IOW, it's not possible to find a line through (-1, 3) that is tangent to the graph of y = x². I apologize for wasting your time with bad advice.

Are you sure that you have posted the problem correctly?

(-1, 3) is inside the parabola! No, there is no line through (-1, 3) tangent to y= x².

 

[ystael]

You don't. That's my point. There aren't any tangent lines to this graph that pass through $$(-1, 3)$$. The question is in error.

 

[Mark44]

Please don't double-post. This is the same question you asked in this thread: https://www.physicsforums.com/showthread.php?t=375109.