dibuje la región encerrada por las curvas dadas.\n$x = 5y^2$, $x = 4 + 4y^2$

Explanation:

Step1: Identify the given curves

The curves are two parabolas opening to the right: $x = 5y^2$ and $x = 4 + 4y^2$.

Step2: Find the intersection points

Set the equations equal to find the $y$-coordinates where the curves meet. $$5y^2 = 4 + 4y^2$$ $$y^2 = 4 \implies y = \pm 2$$

Step3: Determine the $x$-coordinates

Substitute $y = \pm 2$ into either equation to find the $x$-coordinates. $$x = 5(2)^2 = 20$$ The intersection points are $(20, 2)$ and $(20, -2)$.

Step4: Analyze the region

The curve $x = 4 + 4y^2$ is to the right of $x = 5y^2$ for $y \in (-2, 2)$ because at $y=0$, $4 > 0$. The region is bounded by these two parabolas between $y = -2$ and $y = 2$.

Answer:

The region is enclosed by two parabolas opening to the right. The first parabola, $x = 5y^2$, has its vertex at $(0,0)$. The second parabola, $x = 4 + 4y^2$, has its vertex at $(4,0)$. The two curves intersect at the points $(20, 2)$ and $(20, -2)$. To draw it, plot these vertices and intersection points, then shade the area between the curves from $y = -2$ to $y = 2$.