# If a point lies on a conic section, its polar is the tangent through this point to the conic section

I found this fact:

"If a point lies on a conic section, its polar is the tangent through this point to the conic section"

Unfortunately I couldn't find a proof and I have no idea how to prove it

• What's your definition of polar? – Lord Shark the Unknown May 4 at 10:02
• I use the definition from Russian wikipedia (I hope that my translation is correct) "a polar of a point P is a set of points N harmonically conjugate with point P with respect to points M1 and M2 of the intersection of a second-order curve that cross through point P" sorry for mistakes – user670437 May 4 at 10:08
• I have transformed the first "the conic section" into "a conic section". Do you see why ? – Jean Marie May 4 at 21:31

It seems there are problems with the definition. For example, when your conic is an ellipse and $$P$$ lies outside it, the set of harmonic conjugates of P is a segment, not a line. And when $$P$$ belongs to the conic, I don't see how this definition works at all.
For a circle centered at $$O$$ and having radius $$R$$ define the polar of $$A$$ to be the set $$\{B \quad | \quad \vec{OA}\cdot \vec{OB} = R^2 \}$$ It's a line, since scalar product of vectors $$\vec{OA}$$ and $$\vec{OB}$$ (when it's $$\geq 0$$) is equal to the product of |OA| and |OB'|, where $$B'$$ is projection of $$B$$ onto $$OA$$.
When $$X$$ and $$Y$$ lie on a circle, and $$P$$, $$Q$$ are harmonic conjugates with respect to $$X,Y$$, then $$P$$ lies on the polar of $$Q$$. This can be proved in the following way: $$\text{Let } PX:PY = a:b = QX:QY$$ Let's say that $$P$$ lies outside $$XY$$ and $$Q$$ inside it. Then $$\vec{OQ} = \frac{b}{a+b}\vec{OX} + \frac{a}{a+b}\vec{OY}$$ $$\vec{OP} = \frac{-b}{a-b}\vec{OX} + \frac{a}{a-b}\vec{OY}$$ $$\vec{OP}\cdot \vec{OQ} = \frac{1}{a^2-b^2} (b\cdot \vec{OX} + a\cdot\vec{OY}) \cdot (-b\cdot\vec{OX}+a\cdot \vec{OY}) = R^2$$
In this approach your fact is trivial, since if $$A$$ lies on the circle, then $$OA^2 = R^2$$ and so, $$A$$ lies on its own polar. If there is a second point $$B$$ of intersection between the polar of $$A$$ and the circle, then $$\vec{OA}\cdot\vec{OB}=R^2$$, which forces $$\angle AOB$$ to be zero.