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JEE Main20257 Apr 2025Evening ShiftMathematicsHyperbolaActual

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 5 x+9=0 , respectively. If the product of the focal distances of a point ( , 2 5 ) on the hyperbola is p , then 4 p is equal to

Correct answer

0

Step-by-step solution

Equation of hyperbola is ( x ^2 a ^2 - y ^2 ~b ^2 =1 ) Directrix: ( x = -9 5 ) and corresponding foci ((-5,0) ) ( - a e =- 9 5 ) and (- ae =-5 ) ( 9 e^2 5 =5 e= 25 9 = 5 3 a=3 ) ( b ^2= a ^2 ( e ^2-1 )=9 ( 25 9 -1 )=16 ) Hyperbola ( x ^2 9 - y ^2 16 =1 ) (( , 2 5 ) ) lie on it ( ^2 9 - 20 16 =1 ^2= 36 16 9= 81 4 ) Product for distance of ( ( x ₁ y ₁ ) ) from the two foci ( aligned & = (e x₁+a ) |e x₁-a | & =e^2 x₁^2-a^2 aligned ) For (( , 2 5 ) P = 25 9 81 4 -9= 189 4 ) (4 P =189 )

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