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?
Best Answer
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.