Let f(x)=x^2 and let (a,b) be a point in the plane. Under what conditions on (a, b) do the following occur?

Question

a. There are no tangent lines to f(x)=x^2 that pass through (a, b)

b. There is exactly one tangent line to f(x)=x^2 that pass through (a, b)

c. There are two tangent lines to f(x)=x^2 that pass through (a, b)

Answer 1

To determine this, we can find the equation of the tangent line to the curve at an arbitrary point P: (x₁ , y₁) on the parabola. As you will see, we will be able to write this equation solely in terms of the parameter x₁:

Given P on the curve y =x² , we know that y₁ = x₁² so P: (x₁ , x₁²). We also know the slope of the tangent line is given by y^' = 2x , so the slope of the tangent line thru P will be m = 2x₁.

Using pt-slope form of the line, we get the equation as y - x₁² = 2x₁(x - x₁). Here y and x are the true variables.

If the pt (a , b) lies on this tangent line, it satisfies the tangent line equation above, so we substitute y= b, x = a:

b - x₁² = 2x₁ (a - x₁). Distributing and gathering all terms on the left-hand side gives the following ...

x₁² - 2ax₁ + b = 0 which we can interpret as a quadratic equation in the variable x₁. The discriminant is ...

4a² - 4b which we set > 0 for 2 solutions, = 0 for 1 solution only, and < 0 for no solutions.

Two solutions: 4a² - 4b > 0. a² > b which corresponds to the region in the plane consisting of all points below the parabola (in other words, all the points lying in the exterior of the curve).

Similar logic reveals that points on the parabola have only one tangent line that passes through them, while any points above the curve y = x² have no tangent lines that intersect them.

Answer 2

One thing to note about the function is that it has symmetry.

f(x) = x² so f'(x) = 2x

If you sketch the curve, you note that all tangent lines pass below the curve as you move away from the tangent point.

From this you can determine the three regions.