If x=3 t and y=4 t , then d^2 y d x^2 at the point (x₀, y₀ )= ( 3 2 2 , 2 2 ) , is

Step-by-step solution

We have, aligned & x=3 t and y=4 t & x^2 9 + y^2 16 = ^2 t+ ^2 t & x^2 9 + y^2 16 =1 & 2 x 9 + 2 y 16 d y d x =0 & d y d x = -16 9 x y & d^2 y d x^2 = -16 9 [ 1 y-x d y d x y^2 ] & d^2 y d x^2 = -16 9 [ y-x- ( -16 9 x y ) y^2 ] & = -16 9 [ 9 y^2+16 x^2 9 y^3 ]= -16 9 144 9 y^3 [ x^2 9 + y^2 16 =1 ] & = -16 144 81 (2 2 )^3 [ (x₀, y₀ )= ( 3 2 2 , 2 2 ) ] & = -16 144 81 16 2 = -144 81 2 = -16 9 2 = -8 2 9 & aligned