Answer:
x=2at/(1+t²)
dx/dt =[(1+t²)(2at)'-(2at)(1+t²)']/(1+t²)²
dx/dt=[(1+t²)(2a)-(2at)(2t)]/(1+t²)²
dx/dt=[2a+2at²-4at²]/(1+t²)²
dx/dt=[2a-2at²]/(1+t²)²
y=[b(1-t²)]/(1+t²)
dy/dt=[(1+t²)(b(1-t²))'-b(1-t²)(1+t²)']/(1+t²)²
dy/dt=[((1+t²)(-2bt)-b(1-t²)(2t)]/(1+t²)²
dy/dt=(-2bt-2bt³-2bt+2bt³)/(1+t²)²
dy/dt=(-4bt)/(1+t²)²
dy/dx=(dy/dt)/(dx/dt)
dy/dx=[(-4bt)/(1+t²)²]/[(2a-2at²)/(1+t²)²]
Cancelling common denominator
dy/dx=(-4bt)/(2a-2at²)
dy/dx=(-2bt)/(a-at²)
dy/dx=(2bt)/(at²-a)
Differentiating the above equation
y''=d²y/dx²=D[(2bt)/(at²-a)]
y''=[(at²-a)D(2bt)-(2bt)D(at²-a)]/(at²-a)²
y''=[(at²-a)(2b)(dt/dx)-(2bt)(2at)(dt/dx)]/(at²-a)²
dt/dx=(1+t²)²/(2a-2at²)
y''=[(at²-a)(2b)[(1+t²)²/(2a(1-t²))]-
(4abt²)[(1+t²)²/(2a-2at²)]/(at²-a)²
y''=[(-b)(1+t²)²-(2bt²)[(1+t²)²/(1-t²)]/(at²-a)²
y''=[(-b)(1+t²)²(1-t²)-(2bt²)(1+t²)²]/[(1-t²)(at²-a)²]
y''=[b(1+t²)²(t²-1)-2bt²(1+t²)²]/[(1-t²)(at²-a)²]
Step-by-step explanation:
