Angles subtended by tangents from a point on the focus of a conic section

Question

I know that the tangents from a point to a conic section subtend equal angles on the focus.

However, I have mostly studied conic sections from the perspective of coordinate geometry, so even when there are properties common to all conic, I have to prove them separately for parabola, ellipse, hyperbola, and circle.

The only common denominator between these figures I know of is that they are the locus of points which have a constant ratio of distance from a fixed line and a fixed point. However, I haven't been able to use that much to my aid, having hardly any experience in dealing with these figures in such a way.

Observing the apparent simplicity of the result, is there a simple proof for the theorem?

P.S.: I would be thankful if someone could suggest resources that deal with such results about conic sections, especially if they use synthetic geometry.

Answer 1

Here's the theorem you want to prove:

If two tangents are drawn to a conic from an external point, then they subtend equal angles at the focus. In the case of a hyperbola the points of contact must be on the same branch of the curve, otherwise the subtended angles are supplementary.

I'll give a proof for the same-branch case. The other case can be proved in an analogous way.

Let $OQ$, $OQ'$ be two tangents from point $O$ to a conic with focus $S$ (see figure below). Let $Z$ be the intersection point of line $OQ$ with the directrix related to $S$: we have then $SZ\perp SQ$ (see here for a proof). Drop from $O$ the perpendiculars $OU$ to $SQ$ and $OI$ to the directrix and let $M$ be the projection of $Q$ on the directrix.

From similar triangles we have then:

$$ SU:SQ=ZO:ZQ=OI:QM, $$ that is: $$ SU={SQ\over QM}OI=e\,OI, $$ where $e$ is the eccentricity of the conic. Considering tangent $OQ'$ we can analogously prove $SU'=e\,OI$, where $U'$ is the projection of $O$ on $SQ'$. Hence $SU=SU'$ and triangles $OSU$, $OSU'$ are congruent. It follows that $\angle OSQ=\angle OSQ'$, Q.E.D.