Finding tangent lines that pass through given points

[methionine]

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) = nx^n-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)

 

[SammyS]

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) = nx^n-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 !

 

[methionine]

I'm sorry, what do you mean? "...and y = x³

 

[SammyS]

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).

 

[hotvette]

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.

 

[methionine]

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!