Given:
Vertices:
[tex](5,0),(-5,0)[/tex]
Co-vertices:
[tex](0,4)(0,-4)[/tex]
To find: The ellipse equation
Explanation:
The general form of ellipse equation is,
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]
Where (h, k) is the center and a and b are the length of the semi-major and semi-minor axis.
Here,
[tex](h,k)=(0,0)[/tex]
The length of the major axis is the distance between the vertices.
[tex]\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt[]{(0-0)^2+(-5_{}-5)^2} \\ =\sqrt[]{10^2} \\ =10\text{ units} \end{gathered}[/tex]
The length of the minor axis is the distance between the co-vertices.
[tex]\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt[]{(-4-4)^2+(0-0)^2} \\ =\sqrt[]{8^2} \\ =8\text{ units} \end{gathered}[/tex]
Therefore,
The length of the semi-major axis is 5
The length of the semi-minor axis is 4
So, the standard form of an ellipse equation becomes,
[tex]\begin{gathered} \frac{(x-0)^2}{5^2}+\frac{(y-0)^2}{4^2}=1 \\ \frac{x^2}{5^2}+\frac{y^2}{4^2}=1 \end{gathered}[/tex]
Final answer:
The standard form of an ellipse equation becomes,
[tex]\frac{x^2}{5^2}+\frac{y^2}{4^2}=1[/tex]
