Question

graph each equation.

1. \\(dfrac{x^2}{4} + dfrac{y^2}{9} = 1\\)

Explanation:

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 \((0,0)\). The vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
Since \(a=\sqrt{9} = 3\) and \(b=\sqrt{4}=2\), the vertices are \((0, 3)\) and \((0,- 3)\), and the co - vertices are \((2,0)\) and \((-2,0)\).

Step3: Plot the points and draw the ellipse

Plot the center \((0,0)\), vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\) on the coordinate plane. Then, draw a smooth curve connecting these points to form the ellipse. The major axis is along the \(y\) - axis with length \(2a = 6\) and the minor axis is along the \(x\) - axis with length \(2b=4\).

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with center at the origin, vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. If we were to describe the key points for graphing: )

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\). To draw it, plot these points and sketch a smooth curve through them.