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.
Best Answer
Here's the theorem you want to prove:
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.