Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be (-5, 0) and 5x + 9 = 0, respectively. If the product of the focal distances of a point ( , 2 5) on the hyperbola is p, then 4p is equal to .
Answer
189
Solution
Hyperbola: x2a2 - y2b2 = 1. Focus (-ae, 0) = (-5, 0) and directrix x = - ae = - 95.
ae = 5, ae = 95
a2 = 9, a = 3, e = 53
b2 = a2(e2 - 1) = 9 ( 259 - 1 ) = 16
Point ( , 2 5) on x29 - y216 = 1:
29 = 1 + 2016 = 94 | | = 92
Focal distances: e| | + a and e| | - a, with e| | = 53 92 = 152
152 + 3 = 212 and 152 - 3 = 92
p = 212 92 = 1894
4p = 189
Answer: 189