Question

Can someone help with me with (bi)

Answer 1

Hello,


`x=a*cos(t)==\ \textgreater \ (dx)/(dt) =-a*sin(t)\\ y=a*sin(t)==\ \textgreater \ (dy)/(dt)=a*cos(t)\\ (dy)/(dx) =- (1)/(tan(t)) \\\\ x= (a)/(2) ==\ \textgreater \ \\\\ cos(t)= (1)/(2) ==\ \textgreater \ t= (\pi)/(3) ==\ \textgreater \ y=a* (√(3))/(2) \\\\ y'=- (√(3))/(3) \\\\ \textrm{Equation of the tangent :}\\ y-a*(√(3))/(2)=(x-(a)/(2) )*(- (√(3))/(3))\\\\ y=- ( √(3) )/(3) *x+2*a*(√(3))/(3)\\\\ \boxed{3y= √(3) *(2a-x)}`