7. For the hyperbola a2x2−b2y2=1, the right focus is F. The line l:y=kx+d does not pass through point F, and intersects the right branch of the hyperbola at points P and Q. If the external angle bisector of ∠PFQ intersects l at point A, then the x-coordinate of point A is
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Solution
7. a2+b2a2.
Draw a perpendicular line h from point A to the x-axis. Let the projections of P and Q on line h be P′ and Q′, respectively. Then QQ′PP′=AQAP=FQFP, which means PP′PF=QQ′QF. Therefore, line h is the right directrix of the hyperbola. Hence, the x-coordinate of point A is a2+b2a2.
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