AP EAMCET202318 May 2023Morning ShiftMathematicsHyperbolaActual
Let the points P ₁ ( 4 ), P ₂ ( 3 4 ), P ₃ ( 5 4 ) and P ₄ ( 7 4 ) given in parametric form, lie on the hyperbola x^2 9 - y^2 16 =1 . Then these four points in that order form
Options
- Aa rectangle
- Ba square
- Ca parallelogram
- Da rhombus
Correct answer
A. a rectangle
Step-by-step solution
Since, x^2 9 - y^2 16 =1 ...(i) a=3, b=4 Parametric equation for hyperbola (i) is . x=a , y=b x=3 , y=4 aligned & P₁ ( 4 )= (3 4 , 4 4 )=(3 2 , 4) & P₂ ( 3 4 )= (3 3 4 , 4 3 4 )=(-3 2 ,-4) & P₃ ( 5 4 )= (3 5 4 , 4 5 4 )=(-3 2 , 4) & P₄ ( 7 4 )= (3 7 4 , 4 7 4 )=(3 2 ,-4) aligned P₁ P₃=P₂ P₄= 72+0 = 72 aligned & P₂ P₃=P₄ P₁= 0+16 =4 & P₁ P₂=P₃ P₄= 18+64 = 82 aligned P ₁ P ₂ P ₃ P ₄ is a rectangle.