Number of tangent lines that can be drawn to $y=x^2$ from any arbitary point on $xy$- plane

Question

I made up this question but don't know how to solve it!

Consider $xy$- plane and the graph of $y=x^2$ on it. How many tangent lines can be drawn from any arbitrary point (say $(x_0,y_o)$) to the graph?

Intuitively from points of the graph ( i.e $(x_0,x_0^2)$) only one tangent can be drawn to the graph and from the points above $y=x^2$ (i.e $(x_0,y_0)$ where $x_0^2<y_0$) no tangent line can be drawn. For other points I think it is possible to draw two tangent lines to $y=x^2$.

I'm wondering is my intuition right? And how to prove it mathematically?

Answer 1

You are correct that a point on the parabola has exactly one tangent line.

So starting with a point $(x_0, y_0)$ not on the parabola, let's suppose there is a line through $(x_0, y_0)$ and tangent to the parabola $y=x^2$. Call the point of tangency $(x_1, y_1)$. Then $y_1=x_1^2$.

There are no vertical lines tangent to the parabola, so we can look at just lines which have a real slope $m$. We know the slope $m$ in two different ways:

$$ 2x_1 = m = \frac{y_1-y_0}{x_1-x_0} $$

I'll leave details of the next steps up to you: Treating $x_0$ and $y_0$ as constants and solving this equation for $x_1$ will involve a quadratic equation. Apply what you know about the number of real solutions to a quadratic equation, and the result should end up depending on whether the point $(x_0,y_0)$ is above ($y_0>x_0^2$) or below ($y_0<x_0^2$) the parabola. Finally, point out why we know in the case with two solutions that the two possible values for $x_0$ don't give the same one line.

Answer 2

If you are familiar with calculus, we can solve for the answer in one swoop. The tangent line to the parabola $f(x)=x^2$ going through the point $(t,t^2)$ is,

$$y=f'(t)(x-t)+f(t)$$ $$y=2t(x-t)+t^2$$

Now since we know this will be a tangent line, we can now fix $(x,y)$ and reverse the situation to solve for what values of $t$ we get by the quadratic formula,

$$t^2-2xt+y=0$$ $$t=x \pm \sqrt{x^2-y}$$

Notice we didn't really need to explicitly find $t$ to answer your question, we just care about the discriminant, $x^2-y$. Now we just check how the position of the point affects the discriminant, which in turn tells us how many tangent lines we have.

When $y<x^2$, so a point below the parabola, then we have that the discriminant is positive and so we have two solutions. When $y=x^2$ we have that the discriminant is $0$, giving us exactly $1$ solution. When $y>x^2$ we have negative discriminant, which finally tells us we have $0$ solutions.