QUESTION IMAGE
Question
graph each equation.
- \\(\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1\\)
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 it is a vertical ellipse).
Step2: Find the vertices and co - vertices
For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the center is at \((0,0)\) (since there are no shifts in \(x\) or \(y\) in the equation).
- The length of the semi - major axis \(a=\sqrt{9}=3\), so the vertices are at \((0, a)=(0, 3)\) and \((0,-a)=(0, - 3)\).
- The length of the semi - minor axis \(b=\sqrt{4} = 2\), so the co - vertices are at \((b,0)=(2,0)\) and \((-b,0)=(-2,0)\).
Step3: Plot the points and draw the ellipse
- Plot the center \((0,0)\).
- Plot the vertices \((0,3)\) and \((0, - 3)\) (these are the top and bottom most points of the ellipse).
- Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the right and left most points of the ellipse).
- Then, sketch a smooth curve connecting these points to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.
To graph the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):
- Center: \((0,0)\)
- Vertices: \((0,3)\), \((0, - 3)\)
- Co - vertices: \((2,0)\), \((-2,0)\)
After plotting these points, draw a smooth elliptical curve passing through them. The major axis is along the \(y\) - axis with length \(2a = 6\) and the minor axis is along the \(x\) - axis with length \(2b=4\).
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The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\). The ellipse is symmetric about the \(x\) - axis and \(y\) - axis, with the major axis along the \(y\) - axis (length 6) and the minor axis along the \(x\) - axis (length 4). When plotted on the given coordinate grid, we connect the points \((0,3)\), \((2,0)\), \((0, - 3)\), \((-2,0)\) with a smooth elliptical curve.