Question

determine the equation in standard form of the ellipse that satisfies the given conditions
vertices at (2, -8), (2, 2); foci at (2, -7), (2, 1)

Explanation:

Step1: Determine the center of the ellipse

The center \((h,k)\) of the ellipse is the mid - point of the line segment joining the vertices (or foci).
For two points \((x_1,y_1)\) and \((x_2,y_2)\), the mid - point formula is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
For vertices \((2,-8)\) and \((2,2)\) (or foci \((2,-7)\) and \((2,1)\)), \(h=\frac{2 + 2}{2}=2\), \(k=\frac{-8 + 2}{2}=\frac{-7+1}{2}=-3\). So the center is \((h,k)=(2,-3)\).

Step2: Find the values of \(a\) and \(c\)

The distance between the center \((2,-3)\) and a vertex \((2,-8)\) (or \((2,2)\)): \(a=\vert y - k\vert\). Using the vertex \((2,2)\), \(a=\vert2-(-3)\vert = 5\).
The distance between the center \((2,-3)\) and a focus \((2,-7)\) (or \((2,1)\)): \(c=\vert y - k\vert\). Using the focus \((2,1)\), \(c=\vert1-(-3)\vert = 4\).

Step3: Calculate \(b^{2}\)

We know the relationship \(c^{2}=a^{2}-b^{2}\). So \(b^{2}=a^{2}-c^{2}\).
Substitute \(a = 5\) and \(c = 4\), then \(b^{2}=25 - 16=9\).

Step4: Write the standard form of the ellipse equation

Since the \(x\) - coordinates of vertices and foci are the same, the major axis is vertical.
The standard form of an ellipse with center \((h,k)\), major axis vertical is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\).
Substitute \(h = 2\), \(k=-3\), \(a^{2}=25\), \(b^{2}=9\) into the formula: \(\frac{(x - 2)^{2}}{9}+\frac{(y + 3)^{2}}{25}=1\).

Answer:

\(\frac{(x - 2)^{2}}{9}+\frac{(y + 3)^{2}}{25}=1\)