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
Step1: Identify the conic section
The equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) 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 the major axis is along the \(y\)-axis).
Step2: Find the vertices and co - vertices
For an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), \(a=\sqrt{9} = 3\) and \(b=\sqrt{4}=2\).
- The vertices (end - points of the major axis) are at \((0,\pm a)=(0,\pm3)\).
- The co - vertices (end - points of the minor axis) are at \((\pm b,0)=(\pm2,0)\).
Step3: Plot the points
Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.
Step4: Draw the ellipse
Connect the plotted points with a smooth curve to form the ellipse. The ellipse will be centered at the origin \((0,0)\), with a vertical major axis of length \(2a = 6\) and a horizontal minor axis of length \(2b=4\).
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To graph the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):
- Plot the vertices \((0, 3)\) and \((0, - 3)\) (on the \(y\) - axis, 3 units above and below the origin).
- Plot the co - vertices \((2, 0)\) and \((-2, 0)\) (on the \(x\) - axis, 2 units to the right and left of the origin).
- Draw a smooth ellipse passing through these four points, centered at the origin, with the major axis along the \(y\) - axis and the minor axis along the \(x\) - axis.