Find two points on the parabola x^2 whose tangents are perpendicular (calculus)

Question

Given the parabola f(x)=x^2, I need to find two points, A(a, f(a)) and B(-a, f(-a)), where their tangents cross and are perpendicular to each other.

Answer 1

The slope of the tangent line at every point (x,f(x)) is given by the derivative f'(x)=2x. Hence, the slope of the tangent line at the point (a,f(a)) equals 2a and the slope of the tangent line at the point (-a,f(-a)) equals -2a. Since we want the two tangent lines to be perpendicular we must have that (2a)(-2a)=-1 which implies that -4a^2=-1 or a^2=1/4 which is the same as (a=1/2 or a=-1/2). Thus, the points with perpendicular tangent lines are (1/2,1/4) and (-1/2,1/4).