The problem is related to direct variation. It is given that $y$ varies directly as the square of $x$. When $y = 4$ and $x = 1$, we need to find the relationship between $y$ and $x$.

1. Problem Description

The problem is related to direct variation. It is given that $y$ varies directly as the square of $x$. When $y = 4$ and $x = 1$, we need to find the relationship between $y$ and $x$.

2. Solution Steps

Since $y$ varies directly as the square of $x$, we can write the relationship as:

$y = kx^2$

where $k$ is the constant of proportionality.

We are given that $y = 4$ when $x = 1$. Substituting these values into the equation, we get:

$4 = k(1)^2$

$4 = k(1)$

$k = 4$

Now that we have the value of $k$, we can write the equation as:

$y = 4x^2$

3. Final Answer

The relationship between $y$ and $x$ is $y = 4x^2$.