# Relation between geometric and analytic definition of tangent.

Basically, I want to show that, if a line intersects a smooth curve exactly once, and the curve is on exactly one side of the line, then the line is tangent to the curve where it intersects it. I would be happy if someone can show it for circle, ellipse, parabola and hyperbola. I recently shifted from applying coordinate Geometry on conic sections to euclidean geometry . I can intuitively see that the above is true, but I want a rigorous proof.

• What is your definition of “tangent?” – amd Apr 7 '18 at 3:26

Consider a smooth curve $f(x)$ and a line $g(x)$ and let their point of intersection be $(a,b)$

First, suppose $f(x) \leq g(x)$, satisfying equality only at $a$.

Suppose that $f'(a) \neq g'(a)$. Without loss of generality, suppose $f'(a) > g'(a)$. Then, since $f$ is smooth, we have, by definition of the derivative (or slope of tangent line at $a$), we have for $\lim_{\epsilon \to 0}$

$$f(a + \epsilon) = f(a) + \epsilon f'(a)$$ $$g(a + \epsilon) = g(a) + \epsilon g'(a)$$

Subtracting the bottom from the top yields $$f(a + \epsilon) - g(a + \epsilon) = \epsilon(f'(a) - g'(a))$$ But $f'(a) > g'(a)$, so $f(a + \epsilon) > g(a + \epsilon)$, which contradicts our hypothesis that $f$ stays on one side of the line, $g$, i.e, $f(x) \leq g(x)$.

Notice that we can choose $f'(a) < g'(a)$ and then consider $f(a - \epsilon)$ instead.

In either case, we arrive at $f'(a) = g'(a)$.

The main point is that if the lines weren't tangent, at least of the conditions must be false.

• What if the curve is not a function or $f'(a)\to\infty$? I guess the first problem can be solved by parameterizing f. I can intuitively understand vertical tangents, but how to deal with them rigorously. – Devansh Sehta Apr 7 '18 at 11:37
• You can certainly consider the parametrization of a curve where $lim_{x \to a} f'(x)$ is unbounded, I'm not too sure what more rigour is required. – jaslibra Apr 7 '18 at 12:20

$f(x)=ax^2+bx+c$ intesects a line $g(x)=mx+k$ only if $mx+k=ax^2+bx+c$,
or $ax^2+(b-m)x+(c-k)=0$ and if there is only one root of this, then it forces that $(b-m)^2-4a(c-k)=0$ . In other words:
$m^2-2bm+(b^2-4ac+4ak)=0$ and solving for $m$ (using the quadratic formula) we see that:
$m=\frac{2b \pm \sqrt{4b^2-4b^2+16ac-16ak}}{2}=\frac{2b \pm 4\sqrt{ac-ak}}{2}=b \pm2\sqrt{a}\cdot\sqrt{c-k}$
but this is not enough to deduce that $m=2ax+b$.