QUESTION IMAGE
Question
graph each equation.
- \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\)
graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, origin at (0,0)
Step1: Identify the ellipse standard form
The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a^{2}=9\) and \(b^{2}=4\), and \(a > b\), so it's a vertical ellipse). Here, \(a=\sqrt{9} = 3\) and \(b=\sqrt{4}=2\).
Step2: Find the vertices and co - vertices
- For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\), the vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
- Vertices: When \(x = 0\), from \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\), we have \(\frac{y^{2}}{9}=1\), so \(y^{2}=9\) and \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
- Co - vertices: When \(y = 0\), from \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\), we have \(\frac{x^{2}}{4}=1\), so \(x^{2}=4\) and \(x = \pm2\). So the co - vertices are \((2,0)\) and \((-2,0)\).
Step3: Plot the points and draw the ellipse
- Plot the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane.
- Then, draw a smooth ellipse passing through these four points. The major axis is along the \(y\) - axis with length \(2a=6\) and the minor axis is along the \(x\) - axis with length \(2b = 4\).
To graph the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):
- Locate the vertices at \((0,3)\) and \((0, - 3)\) (on the \(y\) - axis, 3 units above and below the origin).
- Locate the co - vertices at \((2,0)\) and \((-2,0)\) (on the \(x\) - axis, 2 units to the right and left of the origin).
- Draw a smooth, closed curve (ellipse) connecting these four points, with the major axis along the \(y\) - axis and the minor axis along the \(x\) - axis.
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The graph is an ellipse with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\) (the actual drawing should be a smooth ellipse passing through these points as described in the steps).