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      

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 gett+1t+t-1t=2xa+2yb2t=2xa+2ybt=xa+ybt=bx+ayab..................3putting the value of 3 into 12xa=bx+ayab+1bx+ayab2xa=bx+ayab+abbx+ay2xa=(bx+ay)2+(ab)2ab(bx+ay)2bx(bx+ay)=b2x2+a2y2+2bxay+a2b22b2x2+2bxay=b2x2+a2y2+2bxay+a2b22b2x2-b2x2-a2y2=a2b2b2x2-a2y2=a2b2

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Keep posting!!



  • 0
What are you looking for?