यदि (If) `x=a((1+t^(2))/(1-t^(2)))` तथा (and) `y=(2t)/(1-t^(2))`, (find) `(dy)/(dx)` निकालें।

Explanation

यहाँ `x=a((1+t^(2))/(1-t^(2)))=a(-1+(2)/(1-t^(2)))`
`rArr(dx)/(dt)=a(0+2(-1)/((1-t^(2))^(2))*(-2t))=(4at)/((1-t^(2))^(2))`
पुनः `y=(2t)/(1-t^(2))`
`rArr(dy)/(dt)=((1-t^(2)).2-2t(-2t))/((1-t^(2))^(2))=(2(1+t^(2)))/((1-t^(2))^(2))`
`therefore(dy)/(dx)=(dy//dt)/(dx//dt)=(2(1+t^(2)))/((1-t^(2))^(2))xx((1-t^(2))^(2))/(4at)=(1+t^(2))/(2at)`.