Two tangents to $\displaystyle y=x^2-2x-3$ 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.