   SEARCH HOME Math Central Quandaries & Queries  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.