# Easy proof

1. Feb 24, 2005

### danne89

Please help me prove that it doesn't exist any other intersect point than P(x0, x0^2) on the y=2^2 curve, for the tangent line in the very same point.

My work:
y=x^2
l(x)=f¨(x0)(x-x0)+x0^2
= 2x0(x-x0)+x0^2
= 2x0x-2x0^2 + x0^2
= 2x0x - x0^2

2. Feb 24, 2005

### dextercioby

You mean to prove the uniqueness of the tangent to a parabola in one certain (albeit arbitrary) point...?That's trivial.The slope is the same (unique) and they all pass through the same point (namely,the tangence point),therefore,all tangents coincide.

Unless,you meant something else...

Daniel.

3. Feb 24, 2005

### danne89

Hey, isn't the slope the same in two points (symmetry over the y-axis.) ?

Edit: Ohh, my no. It's mirrored...

Last edited: Feb 24, 2005
4. Feb 24, 2005

### dextercioby

Yes,it picks up the minus (due to the cosine,which is negative,once you enter $(\pi/2,\pi)$)...

Daniel.

5. Feb 24, 2005

### mathwonk

in general a line can only intersect a parabola twice at most and tangent intersections count as 2. done.

i.e. when you substiotute the parametrization for the loine into the poarabola equation you get a quadratic which can have only 2 roots, and tangents are exactly those points where the root is a double root.