# I'm confused

1. Oct 15, 2006

### i3uddha

alright so here is the problem.
The parabola y=x**2+3 has two tangents which pass through the point (0,–2). One is tangent to the to the parabola at (A,A2+3) and the other at (–A,A2+3). Find (the positive number) A.

i've attempted this problem several times but haven't been able to find the solution.
could someone tell me step by step how i would start this problem?

2. Oct 15, 2006

### HallsofIvy

Better to use x^2 rather than x**2 (I haven't seen that notation since BASIC). Any line through (0, -2) can be written y= mx- 2. A line tangent to y= x^2+ 3 at (A, A^2+ 3) must have slope m= 2A: y= 2Ax- 2
Determine A so that y= A^2+ 3= 2A(A)- 2= 3A^2- 2.

3. Oct 15, 2006

### i3uddha

how did you get the slope to be 2A?
and what would the answer for A be?

4. Oct 15, 2006

### HallsofIvy

The derivative of y= x^2 is y'= 2x. At x= A, y'= 2A and that is the slope ofthe tangent line.

Surely you can solve A^2+ 3= 3A^2- 2 yourself.

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