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Write an Equation for the Ellipse Graphed in Standard Form

Write an equation for the ellipse graphed in standard form by reading its center, vertices, and co-vertices from the graph.

Subject: Math Algebra Chapter: Functions Topic: Function Transformer Answer included
write an equation for the ellipse graphed in standard form ellipse equation standard form ellipse from graph center of ellipse vertices and co-vertices semi-major axis semi-minor axis conic sections
Accepted answer Answer included

Problem

The keyword “write an equation for the ellipse graphed in standard form” requires reading key points from a graph and writing the ellipse equation in standard form. Since no specific graph is provided, the following representative graph description is used:

The ellipse is centered at the point halfway between its left and right vertices. The vertices are at \((-3,-1)\) and \((7,-1)\), and the co-vertices are at \((2,2)\) and \((2,-4)\).

Standard forms for an ellipse

Horizontal major axis: \[ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \] where the vertices are \((h\pm a,\,k)\) and co-vertices are \((h,\,k\pm b)\).

Vertical major axis: \[ \frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1 \] where the vertices are \((h,\,k\pm a)\) and co-vertices are \((h\pm b,\,k)\).

Step-by-step: write the ellipse equation in standard form

1) Identify the center \((h,k)\)

The center is the midpoint of the two opposite vertices on the major axis. Using \((-3,-1)\) and \((7,-1)\):

\[ h=\frac{-3+7}{2}=2,\qquad k=\frac{-1+(-1)}{2}=-1 \]

So the center is \((h,k)=(2,-1)\).

2) Determine whether the major axis is horizontal or vertical

The vertices \((-3,-1)\) and \((7,-1)\) share the same \(y\)-coordinate, so they lie on a horizontal line. This indicates a horizontal major axis.

3) Find the semi-axis lengths \(a\) and \(b\)

The distance from the center to a vertex is \(a\). From \((2,-1)\) to \((7,-1)\):

\[ a=|7-2|=5 \]

The distance from the center to a co-vertex is \(b\). From \((2,-1)\) to \((2,2)\):

\[ b=|2-(-1)|=3 \]

4) Substitute into the standard form

Since the major axis is horizontal, use \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\). Substitute \(h=2\), \(k=-1\), \(a=5\), \(b=3\):

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

Key points check (quick verification)

Plugging a vertex such as \((7,-1)\) into the equation should satisfy it:

\[ \frac{(7-2)^2}{25}+\frac{(-1+1)^2}{9} =\frac{25}{25}+0 =1 \]

The same check works for \((-3,-1)\), \((2,2)\), and \((2,-4)\).

Summary table (read from the graph)

Feature From the graph Value used in standard form
Center Midpoint of \((-3,-1)\) and \((7,-1)\) \((h,k)=(2,-1)\)
Major axis direction Vertices share the same \(y\) Horizontal
Semi-major axis Distance center → vertex \(a=5\)
Semi-minor axis Distance center → co-vertex \(b=3\)
Ellipse equation Standard form substitution \(\frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1\)

Visualization: ellipse with center, vertices, and co-vertices

x y (2, -1) (-3, -1) (7, -1) (2, 2) (2, -4) a = 5 b = 3
The graph highlights the center \((2,-1)\), vertices \((-3,-1)\) and \((7,-1)\), and co-vertices \((2,2)\) and \((2,-4)\), which determine \(a=5\) and \(b=3\) for the standard-form ellipse equation.

Final answer

The ellipse equation in standard form is \[ \frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1. \]

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