Finding tangent lines that pass through given points

  • Thread starter methionine
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  • #1

Homework Statement


Find the co-ordinates of all points on the curve f(x)= x3 whose tangent lines pass through the point (a,0)


Homework Equations


f '(x) = nxn-1


The Attempt at a Solution


I am really not sure how to attack this question. My initial thoughts are to find f '(x) then, given the points (a,0), create a point-slope equation for a line. But, this would only give me one equation. On the other hand, when I look at the graph of x^3 I can't see how two tangent lines could share a point. I'm not sure how I would find the co-ordinates either.

f '(x) = 3x2

y = 3x2 (x-a)
 

Answers and Replies

  • #2
SammyS
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Homework Statement


Find the co-ordinates of all points on the curve f(x)= x3 whose tangent lines pass through the point (a,0)


Homework Equations


f '(x) = nxn-1


The Attempt at a Solution


I am really not sure how to attack this question. My initial thoughts are to find f '(x) then, given the points (a,0), create a point-slope equation for a line. But, this would only give me one equation. On the other hand, when I look at the graph of x^3 I can't see how two tangent lines could share a point. I'm not sure how I would find the co-ordinates either.

f '(x) = 3x2

y = 3x2 (x-a)
... and y = x3 .

Hello methionine. Welcome to PF !
 
  • #3
I'm sorry, what do you mean? ".....and y = x3
 
  • #4
SammyS
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I'm sorry, what do you mean? ".....and y = x3
Your line passes through the point (x, x3) as well as the point (a, 0).
 
  • #5
hotvette
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On the other hand, when I look at the graph of x^3 I can't see how two tangent lines could share a point.

I believe there are indeed two lines tangent to the curve that go through the point (a,0). Look carefully.
 
  • #6
Hey guys, I spent the last day mulling this question over in my head, and decided to attack it in a slightly different way.

Basically, did what I was doing up to this point, set a point-slope form equation using x and x and x3 and x1 and y1....

y-x3 = 3x2(x-x3)

Now, I know the line has to pass through (a,0) as well, so I plugged those values into X and Y and ended up with two points.

-x3 = 3x2(a-x3)

Kind of an interesting question for me. I hope my work checks out!
 

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