Solve this: 185 A curve has the parametric equation x = a2t (t2 + 1) and y - Maths - Conic Sections

Question

Solve this:

185   A   c u r v e   h a s   t h e   p a
r a m e t r i c   e q u a t i o n   x   =   a 2
t   ( t 2   +   1 )   a n d   y   = b
2 t   ( t 2   -   1 )   t h e n   i t s
  e q u a t i o n   i n   r e c tan g u l a r  
c a r t e s i a n   c o - o r d i n a t e   i s a  
x 2 a 2   +   y 2 b 2   =   14    
                 
                 
              b   x 2
  +   y 2   =   a 2 b 2 c   b 2 x 2  
-   a 2 y 2   =   a 2 b 2      
                 
              d   n o n e
  o f   t h e s e

Sandeep Saurav · Accepted Answer

Dear Student,
The given parametric equations are x=a2t(t2+1) and  y=b2t(t2-1)So,  x=a2t×t2+a2t×1            and           y=b2t×t2-b2t×1 x=at2+a2t           and           y=bt2-b2t x=a2(t+1t)           and           y=b2(t-1t)2xa=t+1t
1           and           2yb=t-1t
2Adding 1 and 2, we get⇒t+1t+t-1t=2xa+2yb⇒2t=2xa+2yb⇒t=xa+yb⇒t=bx+ayab
3putting the value of 3 into 1⇒2xa=bx+ayab+1bx+ayab⇒2xa=bx+ayab+abbx+ay⇒2xa=(bx+ay)2+(ab)2ab(bx+ay)⇒2bx(bx+ay)=b2x2+a2y2+2bxay+a2b2⇒2b2x2+2bxay=b2x2+a2y2+2bxay+a2b2⇒2b2x2-b2x2-a2y2=a2b2⇒b2x2-a2y2=a2b2

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