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 (end - points of the major axis) are at \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are at \((\pm b,0)\).
Given \(a^{2}=9\), then \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\).
Given \(b^{2}=4\), then \(b = 2\), so the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points and draw the ellipse

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0,-3)\) (on the y - axis, 3 units above and below the center).
• Plot the co - vertices \((2,0)\) and \((-2,0)\) (on the x - axis, 2 units to the right and left of the center).
• Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be wider along the y - axis (major axis) and narrower along the x - axis (minor axis).

(Note: Since the question 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 \((\pm2,0)\). If we were to describe the key points for graphing: center \((0,0)\), vertices \((0,3),(0, - 3)\), co - vertices \((2,0),(-2,0)\) and the ellipse passing through these points.)

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 connecting them, with the major axis along the \(y\) - axis and the minor axis along the \(x\) - axis.