Question

The graph of an ellipse is given. Match each graph to its equation.

$$\begin{gathered} \left(A\right)\frac{x^2}{4}+y^2=1 \\ \left(B\right)x^2+\frac{y^2}{4}=1 \\ \left(C\right)\frac{x^2}{16}+\frac{y^2}{4}=1 \\ \left(D\right)\frac{x^2}{4}+\frac{y^2}{16}=1 \end{gathered}$$

Short answer

The correct answer is$\left(B\right)x^2+\frac{y^2}{4}=1$

Step-by-step solution

Step 1. Given Information and Method to find: 

We have to match the equation with its graph.

We know that

The equation of an ellipse with its center in the point (0,0) is given with:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

if the major axis is parallel to the x-axis, or with

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

if the major axis is parallel to the y-axis.

In both cases$a>b$

In this case, from the graph we can see that the major axis is parallel to y-axis since its passes through the vertices which are located on the y-axis.

Step 2. Solution

From the previous step we get that the ellipse has to have an equation of the following form:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

From the graph we can see that the vertices are in the following points $(0,2)and(0,-2)$.

From the fact that any ellipse that has its major axis parallel to the y-axis, has its vertices located in$(0,-a)and(0,a)$

it follows that $a=2$.

Since the height of the ellipse is visible to be $b=1$it follows that the equation of the ellipse is

$$\begin{gathered} \frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \\ \Rightarrow \frac{x^2}{1^2}+\frac{y^2}{2^2}=1 \\ \Rightarrow x^2+\frac{y^2}{4}=1 \end{gathered}$$

Step 3. Conclusion:

The correct answer is$\left(B\right)x^2+\frac{y^2}{4}=1$