conic sectionsconvex-analysisgeometry
In Spivak's chapter on differentiation, he asks the reader to prove that the tangent line to an ellipse or a hyperbola intersect the figure at exactly one point.
How is this done most elegantly? I can crunch the numbers and show that it comes out right, but I have a feeling there's a more elegant way. And why is it (is it?) that a smooth region in $\mathbb{R}^n$ is convex iff for every tangent hyperplane, that hyperplane intersects the set in only one point?
Let the foci be $F_1$ and $F_2$ and let a point of intersection be $P$. The tangent to the ellipse is the external angle bisector of $\angle F_1PF_2$, and the tangent to the hyperbola is the internal angle bisector.
Proving these statements without calculus is onerous. Here's a sketch of a partial proof for the statement about ellipses, to give the main ideas:
Let $F_1,F_2,P$ be as above. Let $b$ be the external bisector of $\angle F_1PF_2$; we want to show that $b$ is tangent to the ellipse. By the definition you've been given, that means we want to show that $b$ intersects the ellipse only at $P$. Consider first a point $Q$ on the other side of $b$ from the foci; I claim that $Q$ is not on the ellipse. To show this, draw $F_2Q$, meeting $b$ at $R$; also reflect $F_2$ in $b$ to obtain its image $F_2'$. Note that $F_1PF_2'$ are collinear because $b$ is the external angle bisector. Then $$ F_1Q+QF_2 = F_1Q + QR + RF_2 = F_1Q + QR + RF_2' > F_1F_2' = F_1P + PF_2' = F_1P + PF_2 $$ So $Q$ is not on the ellipse. I'm skipping some details here, notably why it has to be $>$ and not $\ge$ in the triangle inequality here. After you fill in that detail, and if you believe that the ellipse is a continuous curve, then we know that the ellipse doesn't cross $b$, so it is tangent to the ellipse. Then give a similar analysis for hyperbolas.
(We have not proved that conic sections have only one tangent at each point; I think I proved that once from your definition, but I only remember that it was unpleasant... and maybe my proof was wrong anyway.)
The nice way to prove these statements is with the notion of gradient from multivariable calculus: the direction of the gradient of the scalar field "distance from $F_i$" is obviously a unit vector pointing away from $F_i$; from this and the definition of an ellipse as a level curve of the sum of two such scalar fields it's clear that the normal to the ellipse bisects the desired angle. (On preview, along the lines of achille hui's answer.)
The $125$ and the $100$ in your hyperbola equation should have been swapped: $$\frac{x^2}{125}-\frac{y^2}{100}=1$$
This gives $P$ as $\left( \dfrac{25\sqrt5}3,\dfrac{40}{3} \right)$ and $PA \times PB \,$ yields the given answer of $500$.
Best Answer
Both ellipse and hyperbola are conics, and as such the solution to a quadratic equation. Along any line, there are up to two points where the quadric intersects the line. These two points can either be distinct real points, in which case the curve of the quadric switches side at each such point. Or it can be a pair of complex points which are conjugate to one another in those cases where the line passes by the quadric. Or as a limit between these two cases, the two points of intersection can coincide, in which case the quadric will stay on one side of the line. That last one is exactly your tangential situation. So there you see that you have only one point of intersection for tangents. The core is the quadratic nature of the equation.