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

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.

• Is anything wrong with my answer? Commented Oct 31, 2023 at 19:08
• @Intelligentipauca No, the proof is perfect. This is what I was looking for. Do you have any resources to study conics in such a manner, like a book to study common properties of conics? Commented Nov 3, 2023 at 5:56
• Most old books define conics from focus-directrix property, but they usually treat ellipse, hyperbola and parabola separately. See these for instance: archive.org/details/in.ernet.dli.2015.500964 and archive.org/details/cu31924031271509 Commented Nov 3, 2023 at 7:51

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.
$$SU:SQ=ZO:ZQ=OI:QM,$$ 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.