Two tangents from a point $T$ are drawn to parabola $y^2=4ax$ at $P$ and $Q$. If third tangent at $R$ meets other two tangents at $P'$ and $Q'$ find $$\frac{TP'}{TP}+\frac{TQ'}{TQ}$$
My Try:
Let $P(at_1^2, 2at_1)$ , $Q(at_2^2, 2at_2)$ and $R(at_3^2, 2at_3)$ be three points.
we have Equation of $TP$ as $x-t_1y+at_1^2=0$ and Equation of $TQ$ as $x-t_2y+at_2^2=0$ So Point $T$ is given by $(at_1t_2, a(t_1+t_2))$
Similarly we have equation of Tangent at $R$ as $x-t_3y+at_3^2=0$
hence $P'(at_1t_3, a(t_1+t_3))$ and $Q'(at_2t_3, a(t_2+t_3))$
Now $$TP=a|t_2-t_1|\sqrt{1+t_1^2}$$
Similarly $$PP'=a|t_3-t_1|\sqrt{1+t_1^2}$$
Like wise
$$TQ=a|t_2-t_1|\sqrt{1+t_2^2}$$ and
$$Q'Q=a|t_2-t_3|\sqrt{1+t_2^2}$$
Now $$\frac{TP'}{TP}+\frac{TQ'}{TQ}=\frac{TP-P'P}{TP}+\frac{TQ-Q'Q}{TQ}=2-\left(\frac{|t_3-t_1|}{|t_2-t_1|}+\frac{|t_2-t_3|}{|t_2-t_1|}\right)$$
But how to remove Modulus signs and is there any other geometrical approach?
This is a projective solution. It is ''well know'' that the map $\pi: t_P\to t_Q$ defined with $$ P'\longmapsto Q'$$ is projective transformation (i.e. preserves cross ratio and it is bijective). Since \begin{eqnarray*} Q &\mapsto & T \\ T &\mapsto & P \\ \infty &\mapsto & \infty \\ Q' &\mapsto & P' \end{eqnarray*} we have \begin{eqnarray*} (Q,Q';T,\infty) &=& (T,P';P,\infty) \\ {\overrightarrow{QT}\over \overrightarrow{TQ'}} &=& {\overrightarrow{TP}\over \overrightarrow{PP'}} \\ {\overrightarrow{TQ'}\over \overrightarrow{QT}} &=& {\overrightarrow{PP'}\over \overrightarrow{TP}} \\ {\overrightarrow{Q'T}\over \overrightarrow{TQ}} +1&=& {\overrightarrow{PP'}\over \overrightarrow{TP}} +1 \\ 1&=& {\overrightarrow{PP'}+\overrightarrow{TP}\over \overrightarrow{TP}} -{\overrightarrow{Q'T}\over \overrightarrow{TQ}} \\ {\overrightarrow{TP'}\over \overrightarrow{TP}} +{\overrightarrow{TQ'}\over \overrightarrow{TQ}} &=& 1\\ \end{eqnarray*}