goodar2006 wrote: From a point $T$ outside of the ellipse, we draw two tangent lines $TP$ and $TQ$ to the ellipse $\epsilon$. Prove that $\frac{TP}{TQ}\ge \frac{b}{a}$. How do we know which is $TP$ and which $TQ?$ I mean, one of $\frac{TP}{TQ}$ or $\frac{TQ}{TP}$ is always greater than $1,$ also greater than $\frac{b}{a}.$ Can you clarify?

Well, it's very clear that when the problem asks you to prove that $\frac{TP}{TQ}\ge \frac{b}{a}$, you may assume WLOG that $TP\le TQ$, because in the other case the problem will be trivial

If $a = b$ the figure is a circle and the tangents from any point are equal, so suppose $a \ne b$. Impose a coordinate system such that the major and minor axes of the ellipse lie on the $x$ and $y$ axes respectively. Also, by homogeneity we can let $b = 1$ (whence $a > 1$). Let $T$ have coordinates $(p, q)$, and let the points of tangency $P$ and $Q$ have coordinates $(x_1, y_1)$ and $(x_2, y_2)$. Consider a dilation of the plane by factor $\frac{1}{a}$ with respect to the $y$ axis. This sends $T$, $P$, $Q$ to $T'(\frac{p}{a}, q)$, $P'(\frac{x_1}{a}, y_1)$, $Q'(\frac{x_2}{a}, y_2)$. It also sends $\epsilon$ to a circle $\epsilon'$. This means the tangents from $T'$ to $\epsilon'$, that is, $T'P'$ and $T'Q'$, are equal in length. Therefore \[\sqrt{(\frac{p - x_1}{a})^2 + (q - y_1)^2} = \sqrt{(\frac{p - x_2}{a})^2 + (q - y_2)^2} \; \; \; \; \; \; (1)\] Suppose for the sake of contradiction that $TP < \frac{TQ}{a}$. Then \[aTP = \sqrt{a^2(p - x_1)^2 + a^2(q - y_1)^2} < \sqrt{(p - x_2)^2 + (q - y_2)^2} = TQ\] Since $a > 1$, \[(aTQ')^2 = (p - x_1)^2 + a^2(q - y_1)^2 < a^2(p - x_1)^2 + a^2(q - y_1)^2 = (aTP)^2 < (TQ)^2 \] \[ (p - x_2)^2 + a^2(q - y_2)^2 < (p - x_2)^2 + (q - y_2)^2\] contradicting $a > 1$. Equality holds when $(p, q) = (\pm a, \pm b)$ or when $\epsilon$ is a circle ($a = b$).