To find the equation of the ellipse given the vertex, co-vertex, and center, let's follow these steps:
1. Identify the lengths of the semi-major and semi-minor axes:
- The vertex is at [tex]\((0, -7)\)[/tex]. This point indicates the end of the semi-major axis along the y-axis because the major axis is the longest diameter of the ellipse.
- The co-vertex is at [tex]\((4, 0)\)[/tex]. This point indicates the end of the semi-minor axis along the x-axis.
2. Calculate the lengths of the semi-major and semi-minor axes:
- The semi-major axis length ([tex]\(a\)[/tex]) is the absolute distance from the center to the vertex: [tex]\(a = |-7 - 0| = 7\)[/tex].
- The semi-minor axis length ([tex]\(b\)[/tex]) is the absolute distance from the center to the co-vertex: [tex]\(b = |4 - 0| = 4\)[/tex].
3. Write the standard form of the ellipse equation centered at the origin:
The standard form of an ellipse equation centered at the origin [tex]\((0, 0)\)[/tex] is:
[tex]\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\][/tex]
Since [tex]\(a = 7\)[/tex] and [tex]\(b = 4\)[/tex], we substitute [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex]:
[tex]\[
a^2 = 7^2 = 49
\][/tex]
[tex]\[
b^2 = 4^2 = 16
\][/tex]
4. Substitute these values into the standard form of the ellipse equation:
[tex]\[
\frac{x^2}{49} + \frac{y^2}{16} = 1
\][/tex]
Therefore, the correct equation of the ellipse is:
[tex]\[
\frac{x^2}{49} + \frac{y^2}{16} = 1
\][/tex]
By comparing with the given options, the correct answer is:
[tex]\[
\frac{x^2}{49} + \frac{y^2}{16} = 1
\][/tex]
