Question

find the standard form of the equation of the ellipse and give the location of its foci. type the standard form of the equation. (type an equation. simplify your answer.)

Explanation:

Step1: Identify the center, a, b

The center of the ellipse is at \((h, k) = (0, 0)\) (midpoint of vertices). The length of the major axis (along x - axis) is from \(x=-6\) to \(x = 6\), so \(2a=12\), \(a = 6\)? Wait, no, wait. Wait, the ellipse intersects x - axis at \(x=-6\) and \(x = 6\)? Wait, no, looking at the graph, the ellipse goes from \(x=-6\) to \(x = 6\) (so the distance between the two x - intercepts is \(12\), so \(2a = 12\), \(a=6\))? Wait, no, wait, the y - intercepts: from \(y = 2\) to \(y=-2\), so \(2b = 4\), \(b = 2\). Wait, no, wait, the standard form of an ellipse centered at \((h,k)\) with major axis along x - axis is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), where \(a>b\). Wait, but let's check the vertices. The ellipse has vertices at \((\pm6,0)\) and co - vertices at \((0,\pm2)\). So center \((h,k)=(0,0)\), \(a = 6\), \(b = 2\). Then we need to find \(c\) for the foci, where \(c^2=a^2 - b^2\). But first, the standard form: \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Wait, \(a = 6\)? Wait, no, wait, the distance from center to the x - intercept is \(a\), so from \((0,0)\) to \((6,0)\) is \(6\), so \(a = 6\), and from \((0,0)\) to \((0,2)\) is \(b = 2\). Then \(c=\sqrt{a^2 - b^2}=\sqrt{36 - 4}=\sqrt{32}=4\sqrt{2}\approx5.66\). But wait, maybe I made a mistake. Wait, the ellipse: let's count the grid. Each grid square: from \(x=-8\) to \(x = 8\), y from \(-8\) to \(8\). The ellipse touches x - axis at \(x=-6\) and \(x = 6\) (so length of major axis is \(12\), \(a = 6\)), and y - axis at \(y = 2\) and \(y=-2\) (length of minor axis is \(4\), \(b = 2\)). So center is \((0,0)\). So the standard form is \(\frac{x^2}{6^2}+\frac{y^2}{2^2}=1\)? Wait, no, wait, \(a\) is the semi - major axis. Wait, but if the major axis is along x - axis, then \(a\) is the distance from center to the vertex on x - axis. So \(a = 6\), \(b = 2\). So the standard form is \(\frac{x^2}{36}+\frac{y^2}{4}=1\)? Wait, but let's check again. Wait, maybe I messed up \(a\) and \(b\). Wait, the ellipse: the length along x - axis: from \(x=-6\) to \(x = 6\), so that's the major axis (longer axis), so \(2a=12\), \(a = 6\). The length along y - axis: from \(y=-2\) to \(y = 2\), so \(2b = 4\), \(b = 2\). So the standard form is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1=\frac{x^2}{36}+\frac{y^2}{4}=1\).

Step2: Verify the standard form

The standard form of an ellipse centered at the origin \((0,0)\) with major axis along the x - axis is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where \(a\) is the semi - major axis and \(b\) is the semi - minor axis. Here, \(a = 6\) (distance from center to x - vertex), \(b = 2\) (distance from center to y - vertex). Plugging into the formula, we get \(\frac{x^2}{36}+\frac{y^2}{4}=1\).

Answer:

\(\frac{x^{2}}{36}+\frac{y^{2}}{4}=1\)