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Question

The equation of the reflection of the hyperbola (x4)216(y3)29=1 about the line x+y2=0 is  
  1. (x+3)216(y+4)216=1
  2. (x+2)216(y+1)29=1
  3. (x+1)29(y+2)216=1
  4. (x+1)216(y+2)29=1


Solution

The correct option is C (x+1)29(y+2)216=1
Any point on the hyperbola is (4+4secθ,3+3tanθ)
Reflection of point with respect to the line x+y2=0 is 
h(4+4secθ)1=k(3+3tanθ)1=2(4+4secθ+3+3tanθ2)2
 h=13tanθ and k=24secθ
 (h+1)232(k+2)242=1
 (x+1)29(y+2)216=1

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