Solving Derivatives Problem: Find Point on y=1-x^2

[rabar789]

Hey everyone, I'm a new Calculus student, and this problem on derivatives is giving me lots of trouble. I can't get around it! Please help.

Homework Statement

Find the coordinates of all points on the graph of y=1-x^2 at which the tangent line passes through the point (2,0).

Homework Equations

Possibly anything related to derivatives and slope, such as:
y-y1 = m(x-x1)
The power rule: nx^(n-1)

The Attempt at a Solution

All I know is the answer is: 2 (plus or minus) squareroot(3)
I took the derivative of the original equation in the hopes that I might see where I should be going, but I'm just stuck. I don't see where the point comes in. If someone could tell me the steps of how to get the answer so I could use that knowledge in the future, I'd really appreciate it!

 

[HallsofIvy]

Okay, the derivative of y= 1- x² is y'= -2x and that tells you that the slope of the tangent line, at x= x₀ is -2x₀.

Write the equation of the staight line, with slope -2x₀ that passes through (2.0). If the coordinates of a point at which that is the tangent line are (x₀,y₀), then they must satisfy both that equation and y'sub]0[/sub]
= 1- x₀². That gives you two equations for x and y.

 

[rabar789]

Hang on, I think I'm on the right track:

y-y1 = m(x-x1) --> y = 2x-4

And then make that equal to the derived equation?

2x-4 = 1-x^2 ?

Which should equal

x^2 + 2x - 5 = 0 ?

I put that in the quadratic formula, and I'm getting an answer that's close. Am I doing the math wrong?