# How do I find two lines that pass through points (x, y) and are tangent to the curve?

• May 13th 2011, 05:33 PM
samstark
How do I find two lines that pass through points (x, y) and are tangent to the curve?
For example, lets say the function of the curve is y = x^3

How would you find two lines that pass through points (2, 8) and are tangent to the curve?

Many thanks.
• May 13th 2011, 05:39 PM
Prove It
It's possible that there's only one tangent that goes through that line and is tangent to the curve, but anyway...

Your tangent is a line, so is of the form y = mx + c.

The gradient of this tangent will be the same as the gradient of the curve at a point. The gradient of the curve at any point is given by the derivative.

So differentiate y = x^3 to get the derivative, and substitute this derivative (as a function of x) as m.

Can you get to there? Can you see what to do next?
• May 13th 2011, 05:53 PM
samstark
Quote:

Originally Posted by Prove It
It's possible that there's only one tangent that goes through that line and is tangent to the curve, but anyway...

Your tangent is a line, so is of the form y = mx + c.

The gradient of this tangent will be the same as the gradient of the curve at a point. The gradient of the curve at any point is given by the derivative.

So differentiate y = x^3 to get the derivative, and substitute this derivative (as a function of x) as m.

Can you get to there? Can you see what to do next?

y = x^3
y' = 3x^2

I think this is where I am stuck. I'm not sure how to proceed next :(
• May 13th 2011, 05:56 PM
Prove It
OK, you don't know the point where the tangent will touch your curve, so you are just going to use your derivative as your m value for the moment.

Substituting gives y = (3x^2)x + c, or y = 3x^3 + c

Now use your point to find c...
• May 13th 2011, 06:05 PM
samstark
Am I supposed to substitute this way?

8 = 3(2)^3 + c
8 = 24 + c
c = 8 - 24 or c = -16
• May 13th 2011, 06:07 PM
Prove It
Correct.

That means your tangent line is of the form y = mx - 16.

You are almost there, now you need to use your point to find m.
• May 13th 2011, 06:13 PM
samstark
Quote:

Originally Posted by Prove It
Correct.

That means your tangent line is of the form y = mx - 16.

You are almost there, now you need to use your point to find m.

8 = m(2) - 16
8 = 2m - 16
8 +16 = 2m
24 = 2m
24/2 = m
m = 12

Therefore

y = 12x - 16

Is this correct?
• May 13th 2011, 06:20 PM
Prove It
Looks good to me...
• May 13th 2011, 06:26 PM
samstark
Thanks a lot! :D