# Math Help - Points of tangency.

1. ## Points of tangency.

Two tangents to $y=x^2-2x-3$ pass through (5,3). Find the points of tangency.

2. Originally Posted by Hibijibi
Two tangents to $y=x^2-2x-3$ pass through (5,3). Find the points of tangency.
Let the tangent to the curve be at (a, b). Note that b = a^2 - 2a - 3.

The gradient m of the tangent is found by getting the value of the derivative at x = a:

dy/dx = 2x - 2 therefore m = 2a - 2.

So the equation of the tangent is

y - b = m(x - a)

=> y - a^2 + 2a + 3 = (2a - 2)(x - a)

=> y = (2a - 2)x - 2a^2 + 2a + a^2 - 2a - 3 = (2a - 2)x - a^2 - 3.

But the point (5, 3) lies on this tangent.

Therefore 3 = (2a - 2)(5) - a^2 - 3

=> 3 = 10a - 10 - a^2 - 3

=> a^2 - 10a + 16 = 0

=> (a - 2)(a - 8) = 0

=> a = 2, 8.

Substitute into b = a^2 - 2a - 3 to get the corresponding values of b.

3. Thanks! Really appreciate the time taken to help an uneducated sap such as myself.