When the $\angle BPA$ is greatest that $P$ is point on $x$-axis?

Question

Likewise this picture, the angle $\angle BPA$ isn't greatest when $AP+PB$ is least. Than, when the $\angle BPA$ is greatest?I think that $\tan\alpha \tan\beta$ is minimum when $\alpha=\angle GPB,\beta=\angle APF$. Is this right?

Answer 1

Assuming $A$ and $B$ lie on the same side of the $x$-axis.

Take a point $C$ on the $x$-axis, and draw the circle going through $A, B$ and $C$. If there are points of the $x$-axis inside this circle, then any of those points will make a larger angle than $\angle ACB$. This is a corollary of the inscribed angle theorem, which says that any point on the circumference of the circle (on the same side of $AB$) gives the same angle. Thus for any point $D$ of the circle below the $x$-axis, any point of the $x$-axis inside the triangle $\triangle ADB$ will give a point on the $x$-axis yielding a larger angle than $\angle ACB$.

So what you want is the circle going through $A$ and $B$ which touches the $x$-axis, and the point where this circle touches the $x$-axis is your $P$. If the line $AB$ is parallel to the $x$-axis, then the solution is easy. If not, then find the point $Q$ where $AB$ intersects the $x$-axis. By the power of a point, we have $|PQ|^2 = |AQ|\cdot |BQ|$. This should be enough to locate $P$, except you get two solutions. These two solutions are the two local maxima for the angle $\angle APB$, one on either side of the line $AB$. The one of these which lies "below" the line is the global maximum (unless $AB$ is vertical, in which case the two solutions are, of course, equivalent).

In your example, as far as I can see, we have $A = (-6, 8)$ and $B = (4, 3)$, meaning $Q = (10, 0)$. This gives $$ |AQ|\cdot |BQ| = \sqrt{256 + 64}\cdot\sqrt{36 + 9}\\ = \sqrt{14400} = 120 $$ So $P = (10-\sqrt{120}, 0)$ (while the solution $(10+\sqrt{120}, 0)$ is discarded). By the law of cosines, we get $$ \cos\angle APB = \frac{|AP|^2 + |BP|^2 - |AB|^2}{2|AP|\cdot |BP|}\\ = \frac{(\sqrt{120}- 16)^2 + 8^2 + (\sqrt{120} - 6)^2 + 3^2 - 10^2 - 5^2}{2\sqrt{(\sqrt{120}- 16)^2 + 8^2}\sqrt{(\sqrt{120} - 6)^2 + 3^2}}\\ \approx -0.0182 $$ yielding $\angle APB\approx 91.04^\circ$