By definition, the Standard form of the equation of an ellipse is:
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]
Where the center is:
[tex](h,k)[/tex]
When its center is at the Origin, the equation is:
[tex]\frac{x^2}{a^2}+\frac{y^2^{}}{b^2}=1[/tex]
When:
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]
Where:
[tex]a>b[/tex]
It is horizontal.
And when:
[tex]\frac{x^2}{b^2}+\frac{y^2}{a^2}=1[/tex]
Where:
[tex]a>b[/tex]
It is vertical.
In this case, you know that this ellipse is centered at the Origin, its vertex is:
[tex](4,0)[/tex]
And the co-vertex is at:
[tex](0,3)[/tex]
Analyzing the information given in the exercise, you can idenfity that:
[tex]\begin{gathered} a=4 \\ b=3 \end{gathered}[/tex]
Therefore, you can substitute values into the equation
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]
You get:
[tex]\begin{gathered} \frac{x^2}{4^2}+\frac{y^2}{3^2}^{}=1 \\ \\ \frac{x^2}{16}+\frac{y^2}{9}^{}=1 \end{gathered}[/tex]
The answer is: Last option.
