If y varies inversely as x^2, and y = 2 when x = 4, what is the constant of variation?

Question

please help me solve this. please explain. thank you very much for your help.

Answer 1

Use:
y = k/x^2, Where k = constant, Know: "inverse" y = 1/x

2 = k/4^2

2 = k/(16)

2*16 = k

32 = k

Answer 2

y = k/x^2
2 = k/4^2
2 = k/16
k =32

y = kx^2 would be a direct variation
y = k/x^2 is an inverse variation