AP EAMCET20224 Jul 2022Morning ShiftMathematicsHyperbolaActual
The locus of a variable point whose chord of contact w.r.t. the hyperbola x 2 a 2 - y 2 b 2 = 1 subtends a right angle at the origin is
Options
- Ax 2 4 a 2 - y 2 4 b 2 = 1
- Bx 2 a 2 - y 2 b 2 = x 2 a 4 + y 2 b 4
- Cx a - y b = 1 a 2 + 1 b 2
- Dx 2 a 4 + y 2 b 4 = 1 a 2 - 1 b 2
Correct answer
D. x 2 a 4 + y 2 b 4 = 1 a 2 - 1 b 2
Step-by-step solution
Given, The equation of the given hyperbola is x 2 a 2 - y 2 b 2 = 1           ⋯ 1 Now let h , k be the pole of a chord P Q of the hyperbola Then the straight line P Q is the polar of the point h , k w.r.t. the hyperbola and so its equation is x h a 2 - y k b 2 = 1             ⋯ 2 Now the center of the hyperbola is the origin, let it be point C , Now making 1 homogeneous with the help of 2 , the combined equation of C P and C Q is x 2 a 2 - y 2 b 2 =