If the foci of a hyperbola are same as that of the ellipse x29+ y225=1 and the eccentricity of the hyperbola is 158 times the eccentricity of the ellipse, then the smaller focal distance of the point ( 2, 143 25 ) on the hyperbola, is equal to
- 14 25- 43
- 7 25+ 83
- 7 25- 83
- 14 25- 163
Answer
(3) 7 25- 83
Solution
Ellipse: e= 1- 925= 45, foci (0, 4) (major axis along y-axis)
Hyperbola: y2B2- x2A2=1 with eH= 158 45= 32
BeH=4 ⇒ B= 83; directrices y= BeH= 169
focal distance =eH (distance from the nearer directrix)
PS= 32 ( 143 25- 169 )=7 25- 83
Answer: (3)