A tangent to a curve through a point not on the curve

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Question from Carter, a student:

How does one find the tangent points on a curve, given only the curve's function and the x-intercept of that tangent line?
i.e. Find the point(s) on the curve y = -(x^2) + 1, where the tangent line passes through the point (2, 0).
I know that there will be two such points, one where y is very close to 1, and the other point where y is a large negative number. However, I do not recall how to figure out the tangent line equation given a single intercept and solving to find the tangent points.

Carter,

Suppose that a tangent to the curve y = -x² + 1 at the point P on the curve with coordinates (a, b) passes through (2, 0). Since (a, b) lies on the curve b = -a² + 1. The slope of the tangent at P is the derivative of y = -x² + 1 at x = a which is -2a. Thus the equation of the tangent line is

(y - b) = -2a(x - a)

But (2, 0) is on the tangent so

(0 - b) = -2a(2 - a)

Substitute b = -a² + 1 to get

a² - 1 = -4a + 2a²

Solve for a.

Penny

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.