Question

graph each equation.

1. \\(\frac{x^2}{4} + \frac{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 \(a > b\), the major axis is along the \(y\)-axis).

Step2: Find the vertices and co - vertices

For an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices (end - points of the major axis) are \((0,\pm a)\) and the co - vertices (end - points of the minor axis) are \((\pm b,0)\).
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 vertices \((0,3)\) and \((0, - 3)\) on the \(y\) - axis.
• Plot the co - vertices \((2,0)\) and \((-2,0)\) on the \(x\) - axis.
• Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), symmetric about the \(x\) and \(y\) axes, passing through the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) and forming a closed, oval - shaped curve. (To actually draw it, plot the four points and sketch the ellipse through them.)