# 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

## 1 Answer

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.

My favorite definition is this: do a projective transformation sending your conic to a circle (conics are projective images of circles).

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 $$\vec{OQ} = \frac{b}{a+b}\vec{OX} + \frac{a}{a+b}\vec{OY}$$ $$\vec{OP} = \frac{-b}{a-b}\vec{OX} + \frac{a}{a-b}\vec{OY}$$ $$\vec{OP}\cdot \vec{OQ} = \frac{1}{a^2-b^2} (b\cdot \vec{OX} + a\cdot\vec{OY}) \cdot (-b\cdot\vec{OX}+a\cdot \vec{OY}) = R^2$$

So, for an arbitrary conic, do a projective transformation sending the conic to a circle, define the polar as above, do the inverse projective transformation and call the image of the polar the polar for the conic.

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.