Poles and normals of an ellipse

Question

If the pole of the normal at P on an ellipse x^2/a^2 + y^2/b^2 = 1 lies on the normal at Q, show that the pole of the normal at Q lies on the normal at P.

How do we solve this using coordinate geometry?

Answer 1

This follows from a basic fact about pole-polar relationships: the poles of all of the lines through a point are colinear. So, in particular, if you take an arbitrary line $M$ through the pole of any other arbitrary line $L$, the pole of $M$ lies on $L$.

It’s fairly easy to prove this reciprocity property for this specific ellipse. The pole of the line $lx+my+n=0$ is $\left(-\frac{a^2l}n,-\frac{b^2m}n\right)$. (Ignoring for now the case $n=0$.) An arbitrary line through this point has an equation of the form $$\lambda nx+\mu ny=\lambda a^2l+\mu b^2m.$$ Its pole works out to be $\left(-{\lambda a^2n\over \lambda a^2l+\mu b^2m},-{\mu b^2n\over \lambda a^2l+\mu b^2m}\right).$ It’s easy to see that this point satisfies the equation of the original line. I’ll leave verifying the special cases of poles at infinity to you.

If you use homogeneous coordinate vectors and a matrix representation of the ellipse, you can do all of the above without having to worry about any special cases. In fact, we can prove this pole reciprocity for any nondegenerate conic $Q$. The pole of the line $\mathbf l$ is $Q^{-1}\mathbf l$. An arbitrary line through this point can be written as $\mathbf p\times Q^{-1}\mathbf l$, where $\mathbf p$ is some other point on the line. Its pole is then $$Q^{-1}\left(\mathbf p\times Q^{-1}\mathbf l\right) = \frac1{\det(Q)}\left(Q\mathbf p\times QQ^{-1}\mathbf l\right) = \frac1{\det(Q)}\left(Q\mathbf p\times\mathbf l\right)$$ using a standard cross-product identity, symmetry of $Q$ and $\det(Q)\ne0$. (Incidentally, this says that the pole of the line through two points is the intersection of their polars.) Since $Q\mathbf p\times\mathbf l$ is orthogonal to $\mathbf l$, we have $$\mathbf l^T(Q\mathbf p\times\mathbf l) = 0,$$ so this point lies on $\mathbf l$.