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 (α, 2√5) on the hyperbola is p, then 4p is equal to:
❓ Question Let a hyperbola in standard form have transverse and conjugate axes of lengths 2 a 2a and 2 b 2b , respectively. One focus is at ( − 5 , 0 ) (-5,0) and the corresponding directrix is 5 x + 9 = 0 5x + 9 = 0 . If the product of the focal distances of the point ( α , 2 5 ) on the hyperbola is p p , find 4 p 4p . 🖼️ Question Image ✍️ Short Solution Standard form of horizontal hyperbola: x 2 a 2 − y 2 b 2 = 1 Transverse axis = 2 a 2a , conjugate axis = 2 b 2b . Use focus-directrix definition: For a hyperbola, the eccentricity e = c / a e = c/a and distance from point ( x , y ) (x,y) to focus/focus-directrix property: Distance to focus = e ⋅ distance to directrix . \text{Distance to focus} = e \cdot \text{distance to directrix}. Given one focus ( − 5 , 0 ) (-5,0) and directrix 5 x + 9 = 0 ⟹ x = − 9 5 5x+9=0 \implies x=-\frac{9}{5} , eccentricity: e = c a and c = 5 (since focu...